Solution:

a. If we can write $x_{n}$ in the form $x_{n} = \frac{p_{n}}{q_{n}}$ with $p_{n}$ and $q_{n}$ coprime integers, then $x_{n+1}$ can also be written in the same form. Indeed, if we define the integer $k$ as $k = -\left\lfloor 1 - \frac{1}{x_{n}} \right\rfloor$ (where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$), we have
$$x_{n+1} = 1 - \frac{1}{x_{n}} + k = \frac{(1+k)p_{n} - q_{n}}{p_{n}} = \frac{p_{n+1}}{q_{n+1}}$$
with $p_{n+1}$ and $q_{n+1}$ coprime.
In particular $q_{n+1} \leq p_{n}$ and the inequality is strict whenever $p_{n}$ is not coprime with $(1+k)p_{n} - q_{n}$.
Now, since for $n \geq 1$ we have $0 \leq x_{n} < 1$, necessarily $q_{n} > p_{n} \geq q_{n+1}$. Therefore the sequence of positive integers $q_{m}$ is strictly decreasing, hence it cannot be infinite.

Note. The fraction $\frac{(1+k)p_{n} - q_{n}}{p_{n}}$ is in fact already reduced to lowest terms: indeed $\operatorname{MCD}\left(p_{n}, (1+k)p_{n} - q_{n}\right) = \operatorname{MCD}\left(p_{n}, q_{n}\right) = 1$, where the first equality follows from the well-known property $(a + k b, b) = (a, b)$.

b. The sequence stops precisely when a certain element $x_{n}$ is equal to $0$. Given the formula that defines the sequence, $x_{n} = \{1 - \frac{1}{x_{n-1}}\}$, this tells us that $1 - \frac{1}{x_{n-1}}$ is an integer, say $m$, and hence that $x_{n-1} = \frac{1}{1 - m}$ is a rational number. We can now proceed backwards: if a certain term $x_{h}$ is rational, then from the formula $x_{h} = \{1 - \frac{1}{x_{h-1}}\}$ we obtain that there exists an integer $m_{h}$ such that $x_{h} = 1 - \frac{1}{x_{h-1}} - m_{h}$, from which $x_{h-1} = \frac{1}{1 - m_{h} - x_{h}}$ is also a rational number. Proceeding in this way we obtain (by induction) that $x_{0}$ is also rational, as required.

Alternative solution.
Proving that if $x_{0} > 0$ has a finite sequence, then it is a rational number (that is, $x_{0} = \frac{p}{q}$ for suitable integers $p, q$) is equivalent to proving that if $x_{0} > 0$ is not rational, then its sequence is infinite.
We show that if $x_{n}$ is not rational, then neither is $x_{n+1}$; indeed, if there existed $p, q \in \mathbb{N}$ such that $\frac{p}{q} = x_{n+1} = \{1 - \frac{1}{x_{n}}\}$, we would have $x_{n} = \frac{q}{(1+k)q - p}$ (with $k$ as in the first part of the proof of the exercise), that is, $x_{n}$ would be rational, contrary to the hypothesis.
Therefore, by induction, if $x_{0} > 0$ is not rational, no term of the sequence is rational, so in particular $0$ cannot appear in the sequence.