Solution:

Define a break point to be a positive integer $k$ such that $a[k]$ is a palindrome. Let $c_{1}<c_{2}<\ldots$ be a strictly increasing sequence of all break points. Then $c_{n+2} \leq c_{n+1}+c_{n}$ also holds whenever $c_{n+2}>b_{2}$. Namely, if $b_{j-1}<c_{n+2} \leq b_{j}$, then $c_{n+1} \geq b_{j-1}$ and $c_{n} \geq b_{j-2}$.
For positive integers $p$ and $q$, let $p \sim q$ denote the fact that $a_{p}=a_{q}$.
Let $x, x+y$ and $x+y+z$ be three consecutive break points greater than $b_{2}$. From the condition, we have $z \leq x$. Consider any positive integer $r<x$.
Since $x$ is a break point, $r \sim x-r$.
Since $x+y$ is a break point, $x-r \sim x+y-(x-r)=y+r$.
Since $x+y+z$ is a break point, $y+r \sim x+z-r$.
Hence, $r \sim x+z-r$ for all $r<x$. This implies that $x+z$ is also a break point, which means $y=z$ since we considered consecutive break points. Repeating this argument for the next three break points, we can conclude that there exists an arithmetic sequence of break points with common difference $z$.
Let $(x+n z)_{n}$ be an arithmetic sequence of break points. Consider any positive integer $r$. Let $n$ be a positive integer such that $x+n z>r$.
Since $x+n z$ is a break point, $r \sim x+n z-r$.
Since $x+(n+1) z$ is a break point, $x+n z-r \sim x+(n+1) z-(x+n z-r)=z+r$.
Hence, the sequence $\left(a_{n}\right)_{n}$ is periodic with period $z$, which proves the claim.

Solution:

Similarly to the first solution, $x \sim y$ denotes that $a_{x}=a_{y}$. If we know that $a**$ is palindromic, then $x \sim b-x$ for all $0<x<b$. The notation $a[x, y]$ denotes $\left(a_{x}, a_{x+1}, \ldots, a_{y-1}\right)$. Throughout the proof, all variables denote integers.
We show that the sequence is periodic with $b_{2}-b_{1}$. Note that to show this, it suffices to show that $a\left[b_{n}\right]$ is periodic with $b_{2}-b_{1}$ for all $n$. So the problem statement follows from the following proposition.

Main proposition. $a\left[b_{n+k}\right]$ is periodic with $b_{n+1}-b_{n}$ for all $n, k \geq 1$.
We prove the proposition by induction on $k$.

The case $k=1$. If $0<x<b_{n}$ then by palindromeness of $a\left[b_{n}\right]$ and $a\left[b_{n+1}\right]$, we have $x \sim b_{n}-x \sim b_{n+1}-\left(b_{n}-x\right)=x+\left(b_{n+1}-b_{n}\right)$, so $a\left[b_{n+1}\right]$ is periodic with $b_{n+1}-b_{n}$.

The case $k=2$. We prove the following helpful lemma:

Lemma. Suppose that $t \geq 1$ and $0<x<y$. If $a[x]$ is periodic with $t$, and $a[y]$ is palindromic, and $2 x-y \geq t$, then $a[y]$ is periodic with $t$ too.

Proof of lemma. Since $a[x]$ is periodic with $t$, by palindromeness of $a[y]$ we have that $a[y-x, y]$ is periodic with $t$ too. If the two segments $a[x]$ and $a[y-x, y]$ overlap in at least $t$ elements, this means that all pairs of distance $t$ in $a[y]$ will be contained in at least one of the segments, so $a[y]$ is also periodic with $t$. Here, the overlap is $x-(y-x)=2 x-y \geq t$ indeed.
To use the lemma here, pick $t=b_{n+1}-b_{n}, x=b_{n+1}$ and $y=b_{n+2}$. We already know that $a\left[b_{n+1}\right]$ is periodic with $t$ (by the $k=1$ case), and we have $2 x-y \geq t \Leftrightarrow b_{n+1}+b_{n} \geq b_{n+2}$ indeed, concluding the $k=2$ case.

The case $k \geq 3$. To prove this case, we show that if $a\left[b_{n+k-1}\right]$ is periodic with $b_{n+1}-b_{n}$ and $a\left[b_{n+k}\right]$ is periodic with $b_{n+2}-b_{n+1}$ (both true by the inductive hypothesis) then $a\left[b_{n+k}\right]$ is periodic with $b_{n+1}-b_{n}$ too.
Let $t=b_{n+1}-b_{n}$ and $\Delta=b_{n+2}-b_{n+1}$.
Then we can inductively show that $a\left[b_{n+k-1}+\ell \Delta\right]$ is periodic with $t$ for all $\ell \geq 0$. For $\ell=0$ this is true by the $k=1$ case. If for $\ell \geq 1$, we know that $a\left[b_{n+k-1}+(\ell-1) \Delta\right]$ is periodic with $t$, then by the periodicity of $a\left[b_{n+k}\right]$ with $\Delta$, we have that $a\left[\Delta, b_{n+k-1}+\ell \Delta\right]$ is also periodic with $t$. Also, the two $t$-periodic intervals overlap in at least $t$ elements, as $b_{n+k-1}+(\ell-2) \Delta \geq b_{n+k-1}-\Delta=b_{n+k-1}-b_{n+2}+b_{n+1} \geq b_{n+1}-b_{n} \Leftrightarrow b_{n+k-1}+b_{n} \geq b_{n+2}$, true since from $k \geq 3$, $b_{n+k-1} \geq b_{n+2}$. So $a\left[b_{n+k-1}+\ell \Delta\right]$ is also periodic with $t$.
Now choose $\ell$ so that $d-\Delta \leq b_{n+k-1}+\ell \Delta \leq b_{n+k}$, giving that we have some $0 \leq u \leq \Delta$ with $a\left[b_{n+k}-u\right]$ being $t$-periodic. Now by palindromeness of $a\left[b_{n+k}\right]$, also $\text{a}$ is $t$-periodic, and the two intervals' overlap is at least $t: b_{n+k}-2 u \geq t \Leftrightarrow 2 u \leq b_{n+k}-b_{n+1}+b_{n}$, and $2 u \leq 2 \Delta=2 b_{n+2}-2 b_{n+1} \leq b_{n+k}-b_{n+1}+b_{n} \Leftrightarrow 2 b_{n+2} \leq b_{n+k}+b_{n+1}+b_{n}$, true since $b_{n+2} \leq b_{n+k}$ and $b_{n+2} \leq b_{n+1}+b_{n}$. So $a\left[b_{n+k}\right]$ is $t$-periodic, finishing the proof.

Solution:

We will show that the sequence is periodic with $b_{k}-b_{k-1}$ where $k \geq 2$ is so that $b_{k}-b_{k-1}=\min _{1 \leq n} b_{n}-b_{n-1}$. First, we prove that $b_{k}$ is periodic with $b_{k}-b_{k-1}$ :
$$x \sim b_{k-1}-x \sim b_{k}-b_{k-1}+x \quad \forall x<b_{k-1}$$
where we have used that $b_{k-1}$ and $b_{k}$ are palindromic.
We will use the lemma from the above solution. Let's recall it.

Lemma. Suppose that $t \geq 1$ and $0<x<y$. If $a[x]$ is periodic with $t$, and $a[y]$ is palindromic, and $2 x-y \geq t$, then $a[y]$ is periodic with $t$ too.

Let us apply the lemma repeatedly so that $t=b_{k}-b_{k-1}, x=b_{n-1}$ and $y=b_{n}$ for $n= k+1, k+2, k+3, \ldots$. The two conditions in the first sentence of the lemma are clearly satisfied at such applications, as well as that $a[y]$ is palindromic.
The condition $2 x-y \geq t$ is equivalent to $2 b_{n-1}-b_{n} \geq b_{k}-b_{k-1}$. Since $n \geq k \geq 2$, we know that $b_{n-1}+b_{n-2} \geq b_{n}$. Therefore $b_{n-1}-b_{n} \geq-b_{n-2}$. But since we chose $k$ so that $b_{k}-b_{k-1}$ is smallest, we have that $2 x-y \geq t$ is also satisfied at each application of the lemma.
The only remaining condition is that $a[x]=a\left[b_{n-1}\right]$ is periodic with period $t$. However, since the lemma states that $a[y]=a\left[b_{n}\right]$ is periodic with period $t$, at each application of the lemma we get that this condition is also satisfied for the next application. Therefore we get that $a\left[b_{n}\right]$ is periodic with period $b_{k}-b_{k-1}$ for all $n \geq k$. This finishes the proof.

Solution:

This solution uses the following lemma.

Lemma. If a palindromic, finite sequence $\left(a_{n}\right)_{n=1}^{n=l}$ of length $l$ is periodic with periods $x$ and $y$ with $x<y$ and $x+y \leq l+1$, it is periodic with period $y-x$.

Proof. For $r \leq x-1$, we have $r+y \leq l$ and $x<r+y$, so by periodicity with $y$ and $x$, we have $a_{r}=a_{r+y}=a_{r+y-x}$.
For $x<r \leq l-(y-x)$, we have $1 \leq r-x \leq l-y$, so by the periodicities we have $a_{r}=a_{r-x}=a_{r-x+y}$.
It remains to be proven that $a_{x}=a_{y}$. In case $x+y \leq l$, the first of the two above arguments works for $r=x$ and shows $a_{x}=a_{y}$, proving the lemma.
However, in case $x+y \not l l$, by the condition of the lemma we have $x+y=l+1$. Then by the palindromic condition, $a_{x}=a_{l+1-x}=a_{y}$. This finishes the proof of the lemma.

Comment. Via Euler's algorithm, we may use the lemma repeatedly to give that the sequence is periodic with $\operatorname{gcd}(x, y)$.

As in the previous solution, we use that for all $n \geq 2, a\left[b_{n}\right]$ is periodic with period $b_{n}-b_{n-1}$. Then as $a\left[b_{n}\right]$ is a subsequence for $a\left[b_{n+1}\right]$, we have that $a\left[b_{n}\right]$ is also periodic with period $b_{n+1}-b_{n}$.
Now let's apply the lemma's remark for $a\left[b_{n}\right]$, which we know is palindromic. The two periods are $b_{n}-b_{n-1}$ and $b_{n+1}-b_{n}$. The condition $x+y \leq l+1$ translates to $b_{n+1}-b_{n-1} \leq b_{n}$ (noting that $l=b_{n}-1$ ). However, this is satisfied by the problem statement, so we get that $a\left[b_{n}\right]$ is periodic with period $\operatorname{gcd}\left(b_{n}-b_{n-1}, b_{n+1}-b_{n}\right)$.
Recalling that $a\left[b_{n+1}\right]$ is periodic with period $b_{n+1}-b_{n}$, which is a multiple of the period we got for $a\left[b_{n}\right]$. Since $b_{n+1}-b_{n} \leq b_{n-1} \leq b_{n}-1$, an entire larger period is contained in $a\left[b_{n}\right]$. Then each larger period in $a\left[b_{n+1}\right]$ consists of smaller periods of length $\operatorname{gcd}\left(b_{n}-b_{n-1}, b_{n+1}-b_{n}\right)$ from $a\left[b_{n}\right]$, so $a\left[b_{n+1}\right]$ is also periodic with period $\operatorname{gcd}\left(b_{n}-b_{n-1}, b_{n+1}-b_{n}\right)$.
Then substituting $n+1$ by $n$, $a\left[b_{n}\right]$ is periodic with period $\operatorname{gcd}\left(b_{n-1}-b_{n-2}, b_{n}-b_{n-1}\right)$. So by the remark, it is periodic with period $\operatorname{gcd}\left(b_{n-1}-b_{n-2}, b_{n}-b_{n-1}, b_{n+1}-b_{n}\right)$. By repeatedly using the multiple-period argument, substituting $n+1$ by $n$ and using the remark, we get that $a\left[b_{n}\right]$ is periodic with $\operatorname{gcd}\left(b_{2}-b_{1}, b_{3}-b_{2}, \ldots, b_{n+1}-b_{n}\right)$.
Since $\left(\operatorname{gcd}\left(b_{2}-b_{1}, b_{3}-b_{2}, \ldots, b_{n+1}-b_{n}\right)\right)_{n}$ is a strictly decreasing positive integer sequence, it has a minimum which it achieves at some $n=k$. Then for $p=\operatorname{gcd}\left(b_{2}-b_{1}, b_{3}-b_{2}, \ldots, b_{k+1}-b_{k}\right)$, we have that $a\left[b_{n}\right]$ is periodic with period $p$ for all $n \geq k$. This finishes the proof.

Comment. The lemma is true even if the palindromic condition is dropped. This stronger formulation requires a more in-depth, harder proof for the case $x+y=l+1$, presented below.

Proof. We repeatedly perform the following moves, starting from index $x$. If the current index is at most $l-x$, we increase it by $x$. If we cannot perform this move, and the index is at least $y+1$, we decrease it by $y$. If we cannot perform either move, we stop.
If at some point we cannot perform more steps, the index $r$ we have satisfies $l-x<r<y+1$, so $r=y$. Since at each step the elements of $\left(a_{n}\right)_{n=1}^{n=l}$ at the old and new index are equal due to periodicity, the elements at the first and last indices in our steps are equal, so $a_{x}=a_{y}$.
If we can perform the above steps infinitely many times, there will be an index at which we arrive at least twice. Let $r_{1}$ be the first such index. Then let the subsequent indices we get by the above steps from $r_{1}$ be $\left(r_{n}\right)_{n \geq 0}$. Since $r_{1}=r_{k+1}$ for some index as we arrive at $r_{1}$ twice, we know that taking a step from $r_{k}$ takes it to $r_{1}$, i.e. $r_{1}=r_{k}+x$ or $r_{1}=r_{k}-y$. However, at any index it is clear that we can only arrive by one type of moves, since then $1 \leq r-x \leq l-x$ and $l-x<r+y \leq l$ are both satisfied by $r$ so $1+x \leq r \leq l-y$, contradiction as $l=x+y-1$. So the index preceding $r_{1}$ is also repeated at $r_{k}$ as they cannot be different. So necessarily all indices repeat. But the index preceding the second time we reach $x$ cannot be $0=x-x$ and also cannot be $x+y=l+1$, contradiction. So we can never take infinitely many such steps.