(a) Let us write each term of the sequence as a fraction reduced to lowest terms, $x_{n}=\frac{p_{n}}{q_{n}}$, and consider a new sequence $y_{n}=\max \left\{p_{n}, q_{n}\right\}$. We want to prove that the sequence $y_{n}$ is weakly decreasing, that is, that $y_{n+1} \leq y_{n}$ for every $n \geq 0$. This implies point (a), since it follows that $p_{n}, q_{n} \leq y_{n} \leq y_{0}$, and there is only a finite quantity of distinct terms belonging to the sequence $x_{n}$, because there is only a finite number of rational numbers with numerator and denominator bounded by $y_{0}$. To prove the inequality $y_{n+1} \leq y_{n}$ we distinguish two cases:
- if $p_{n}$ is even, then $x_{n+1}=\frac{\left|p_{n} / 2-q_{n}\right|}{q_{n}}$; we observe that this latter fraction is reduced to lowest terms, because $\operatorname{GCD}\left(p_{n} / 2-q_{n}, q_{n}\right)=\operatorname{GCD}\left(p_{n} / 2, q_{n}\right)=1$, since $p_{n} / q_{n}$ is reduced to lowest terms. We also note that if $a, b$ are two positive numbers we have $|a-b| \leq \max \{a, b\}$. It follows that $y_{n+1}=\max \left\{\left|p_{n} / 2-q_{n}\right|, q_{n}\right\} \leq \max \left\{p_{n} / 2, q_{n}\right\} \leq \max \left\{p_{n}, q_{n}\right\}=y_{n}$.
- if $p_{n}$ is odd, $x_{n+1}=\frac{\left|q_{n}-p_{n}\right|}{p_{n}}$. As before the fraction is reduced to lowest terms, and we have $y_{n+1}=\max \left\{\left|q_{n}-p_{n}\right|, p_{n}\right\} \leq \max \left\{p_{n}, q_{n}\right\}=y_{n}$.

(b) We observe that, since the sequence $x_{n}$ contains only a finite number of distinct terms, at some point one of them will repeat, and the sequence will be periodic from that point onward.
Let us first prove that the values 0 and $2 / 3$ cannot both exist. If the value 0 appears first, for instance $x_{k}=0$, then $x_{k+1}=1, x_{k+2}=0$, and so on, so the value $2 / 3$ cannot appear. Similarly, if the value $2 / 3$ appears first, for instance $x_{m}=2 / 3$, then $x_{m+1}=2 / 3, x_{m+2}=2 / 3$, and so on, so the value 0 cannot appear.
Let us now prove that at least one of the two terms 0 and $2 / 3$ must appear. To this end suppose that 0 does not appear, and let us show that then $2 / 3$ must appear. Let us first observe that, if 0 does not appear, then 1 does not appear either (if $x_{n}=1$ then $x_{n+1}=0$).
Since these are positive integers, the inequality $y_{n+1} \leq y_{n}$ can be a strict inequality only in a finite number of cases, since infinitely many cases of strict inequality would make the value of $y_{n}$ drop below zero. Examining the various cases, we have
(1) $p_{n}>q_{n}$ and $p_{n}$ even: then $y_{n}=p_{n}$ while $y_{n+1}=\max \left\{\left|p_{n} / 2-q_{n}\right|, q_{n}\right\}<p_{n}=y_{n}$;
(2) $p_{n}<q_{n}$ and $p_{n}$ odd: then $y_{n}=q_{n}$ while $y_{n+1}=\max \left\{q_{n}-p_{n}, p_{n}\right\}<q_{n}=y_{n}$;
(3) $p_{n}>q_{n}$ and $p_{n}$ odd;
(4) $p_{n}<q_{n}$ and $p_{n}$ even.
Cases (1) and (2) can occur only a finite number of times, because they present a strict inequality. Hence, from a certain point onward, only cases (3) and (4) can occur. Provided we consider $n$ sufficiently large, we can therefore restrict ourselves to cases (3) and (4).
Suppose we are in case (3). Then $x_{n+1}=\frac{\left|q_{n}-p_{n}\right|}{p_{n}}=\frac{p_{n}-q_{n}}{p_{n}}<1$, so we cannot again be in case (3), and hence $x_{n+1}$ falls into case (4).
Suppose now we are in case (4). Then, $x_{n+1}=\frac{\left|p_{n} / 2-q_{n}\right|}{q_{n}}=\frac{q_{n}-p_{n} / 2}{q_{n}}<1$, so $x_{n+1}$ again falls into case (4).
In conclusion, from a certain point onward we are always in case (4) and, as observed previously, the sequence is periodic. Let $n_{0}$ be an integer such that, for every $n \geq n_{0}$, we fall into case (4) and the sequence is periodic. In particular, there exists an index $k>0$ such that $x_{n_{0}+k}=x_{n_{0}}$. From the equation $x_{n+1}=\frac{\left|p_{n} / 2-q_{n}\right|}{q_{n}}=\frac{q_{n}-p_{n} / 2}{q_{n}}$ it is easy to prove, for example by induction, that $p_{n_{0}+k}=\frac{2^{k}-(-1)^{k}}{3 \cdot 2^{k-1}} q_{n_{0}}+\frac{(-1)^{k}}{2^{k}} p_{n_{0}}$. Finally, knowing that $p_{n_{0}+k}=p_{n_{0}}$, we obtain the equation
$$\frac{2^{k}-(-1)^{k}}{3 \cdot 2^{k-1}} q_{n_{0}}+\frac{(-1)^{k} p_{n_{0}}}{2^{k}}=p_{n_{0}} \Rightarrow \frac{2^{k}-(-1)^{k}}{3 \cdot 2^{k-1}} q_{n_{0}}=p_{n_{0}}\left(\frac{2^{k}-(-1)^{k}}{2^{k}}\right) \Rightarrow \frac{p_{n_{0}}}{q_{n_{0}}}=\frac{2}{3},$$
as desired.