Solution:

a.
We must show that, for $n$ sufficiently large, the number $a b c=2^{x} \cdot 3^{y} \cdot 5^{z}$ divides $c b a \cdot 10^{n}=c b a \cdot 2^{n} \cdot 5^{n}$. We will show that this condition is satisfied with $n=\max \{x, z\}$. Certainly with this choice of $n$ we have that $2^{x} \cdot 5^{z}$ divides $2^{n} \cdot 5^{n}$. It therefore suffices to verify that $3^{y}$ divides $c b a \cdot 10^{n}$, and clearly to do this it is enough to show that $3^{y}$ divides $c b a$. For $y=0$ there is nothing to prove ($3^{0}=1$ divides any number). For $y=1$ we have that 3 divides $a b c$, that is, by the well-known divisibility criterion for 3, that 3 divides $a+b+c$. By the same divisibility criterion we then have that $3=3^{y}$ divides $c b a$ (whose digit sum is again $a+b+c$), which gives the claim. In the case $y=2$ we proceed similarly: the divisibility criterion for 9 guarantees that $a+b+c$ is a multiple of 9, and hence $c b a$ is also a multiple of $3^{y}=9$, as desired.

Second solution.
Let us write more properly $a b c=100 a+10 b+c$ and $c b a=100 c+10 b+a$. We are asked to show that $100 a+10 b+c=2^{x} \cdot 3^{y} \cdot 5^{z}$ divides $(100 c+10 b+a) \cdot 10^{n}$ for $n$ sufficiently large. An integer $k$ divides an integer $m$ if and only if $k$ divides $m-t k$, where $t$ is any integer. In particular, taking $t=10^{n}$, we have that $100 a+10 b+c$ divides $(100 c+10 b+a) \cdot 10^{n}$ if and only if it divides $(100 c+10 b+a) \cdot 10^{n}-(100 a+10 b+c) \cdot 10^{n}=9 \cdot 11 \cdot(c-a) \cdot 10^{n}$. Since by hypothesis $y \leq 2$, it is clear that for $n=\max \{x, z\}$ we obtain the desired divisibility, since $3^{y}$ divides 9 and $2^{x} \cdot 5^{z}$ divides $10^{\max \{x, z\}}=2^{\max \{x, z\}} \cdot 5^{\max \{x, z\}}$.

b.
Let us consider for which $k$ the number $a b c$ divides $c b a \cdot 10^{k}=c b a \cdot 2^{k} \cdot 5^{k}$. Let us write the prime factorization of $a b c$ as $2^{x} \cdot 5^{y} \cdot p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$, where $p_{1}, \ldots, p_{k}$ are primes distinct from each other and distinct from 2 and 5. Similarly, let us write the factorization of $c b a$ as $2^{X} \cdot 5^{Y} \cdot q_{1}^{f_{1}} \cdots q_{r}^{f_{r}}$. The divisibility condition is that every prime appears in the factorization of $a b c$ with an exponent less than or equal to the one with which it appears in $c b a \cdot 10^{k}$. For primes other than 2 and 5, this means that every $p_{i}$ also appears in the factorization of $c b a$, and that the exponent with which it divides $c b a$ is greater than or equal to the one with which it divides $a b c$. This condition does not depend on $k$. For the primes 2 and 5, however, the condition
$$2^{x} \cdot 5^{y} \text{ divides } c b a \cdot 10^{k}=2^{X+k} \cdot 5^{Y+k} \cdot q_{1}^{f_{1}} \cdots q_{r}^{f_{r}}$$
translates into $x \leq X+k$ and $y \leq Y+k$. We then claim that a value of $k$ strictly greater than 9 can never be necessary: indeed $k=10$ could only be necessary if $x \geq 10$ or $y \geq 10$, but in that case $a b c$ would be divisible either by $2^{10}=1024$ or by $5^{10}>1000$, and hence could not be a three-digit number. This shows that the maximum flower cannot exceed 9. On the other hand, taking $a b c=512=2^{9}$, we have that the minimum $k$ for which $a b c$ divides $c b a \cdot 10^{k}=215 \cdot 2^{k} \cdot 5^{k}$ is exactly 9 (the factors of 2 in this last product are exactly $k$). The maximum flower of a three-digit petaloso number is therefore 9.

c.
Suppose for contradiction that 13 divides the petaloso number $a b c$. By what was proved in the previous point we see that 13 also divides $c b a$. Writing these numbers explicitly in terms of their base-10 representation we have $a b c=100 a+10 b+c$ and $c b a=100 c+10 b+a$. If 13 divides both of these numbers, then it must divide their difference, which is $99(c-a)$. Now observe that 13 is a prime number, so the product $99(c-a)$ is divisible by 13 if and only if one of the two factors is. However 99 is not divisible by 13, as is easily checked by direct computation, and $c-a$ is not divisible by 13 since it is the difference of two distinct nonzero digits: indeed we have $0<|c-a| \leq 8$, and hence certainly $c-a$ is not a multiple of 13. We have reached a contradiction, so the original assumption that 13 divided $a b c$ must have been false, which proves the claim.