Solution:

a) Suppose that the inequality $\{a_{n} a_{n-1} \ldots a_{1} x\} > \frac{1}{a_{n+1}}$ holds for finitely many values of $n$. Hence there exists $s$ such that for any $n \geq s$ we have $\{a_{n} a_{n-1} \ldots a_{1} x\} \leq \frac{1}{a_{n+1}}$. Since $\{a_{n} a_{n-1} \ldots a_{1} x\}$ is not a rational number (in particular does not equal $0$) we obtain that $\{a_{n} a_{n-1} \ldots a_{1} x\} < \frac{1}{a_{n+1}}$, i.e. $a_{n+1} \{a_{n} a_{n-1} \ldots a_{1} x\} < 1$. Using that $a_{n+1}$ is an integer we have
$$\{a_{n+1} a_{n} a_{n-1} \ldots a_{1} x\} = \{a_{n+1} \{a_{n} a_{n-1} \ldots a_{1} x\}\} = a_{n+1} \{a_{n} a_{n-1} \ldots a_{1} x\}$$
For any $t > s$ we obtain
$$1 > \{a_{t} a_{t-1} \ldots a_{s} a_{s-1} \ldots a_{1} x\} = a_{t} a_{t-1} \ldots a_{s} \{a_{s-1} \ldots a_{1} x\}$$
a contradiction, since $\lim_{t \rightarrow \infty} a_{t} a_{t-1} \ldots a_{s} = \infty$, but $0 < \{a_{s-1} \ldots a_{1} x\} < 1$.

b) It is clear that if $a_{i} = 1$ for some $i > 1$ then $\{a_{i-1} a_{i-2} \ldots a_{1} x\} > \frac{1}{a_{i}} = 1$ is not true. Suppose that there exists $t$ such that $a_{i} = 2$ for $i > t$. Then $\{2^{p} y\} > \frac{1}{2}$ for $y = a_{t} a_{t-1} \ldots a_{1} x$ and every $p$. Since $y < 1$ and $\sum_{j=1}^{\infty} \frac{1}{2^{j}} = 1$ we conclude that for every $k$ the inequality $c_{k} \leq y < c_{k+1}$ holds true, where $c_{k} = \frac{1}{2} + \frac{1}{2^{2}} + \cdots + \frac{1}{2^{k}}$. Therefore $2^{k} y \in [2^{k} c_{k}, 2^{k} c_{k} + \frac{1}{2})$, a contradiction to $\{2^{k} y\} > \frac{1}{2}$.

We shall prove that if $\{a_{n}\}_{n=1}^{\infty}$ is a sequence for which $a_{i} > 1$ for all $i > 1$ and the inequality $a_{i} > 2$ holds true for infinitely many values of $i$, then there exist infinitely many $x \in (0,1)$ such that $x_{n} > \frac{1}{a_{n+1}}$. Set
$$x = \frac{b_{1}}{a_{1}} + \frac{b_{2}}{a_{1} a_{2}} + \frac{b_{3}}{a_{1} a_{2} a_{3}} + \cdots$$
where $b_{1} \leq a_{1} - 1$ and $1 \leq b_{i} \leq a_{i} - 1$ for $i > 1$ and infinitely many of the latter inequalities are strict. Then
$$\begin{aligned} x & = \frac{b_{1}}{a_{1}} + \frac{b_{2}}{a_{1} a_{2}} + \frac{b_{3}}{a_{1} a_{2} a_{3}} + \cdots \\ & < \frac{a_{1} - 1}{a_{1}} + \frac{a_{2} - 1}{a_{1} a_{2}} + \frac{a_{3} - 1}{a_{1} a_{2} a_{3}} + \cdots \\ & = 1 - \frac{1}{a_{1}} + \frac{1}{a_{1}} - \frac{1}{a_{1} a_{2}} + \frac{1}{a_{1} a_{2}} - \cdots = 1 \end{aligned}$$
The numbers of this type are infinitely many and we have as above that
$$\frac{b_{n+1}}{a_{n+1}} + \frac{b_{n+2}}{a_{n+1} a_{n+2}} + \cdots < 1$$
Therefore
$$x_{n} = \frac{b_{n+1}}{a_{n+1}} + \frac{b_{n+2}}{a_{n+1} a_{n+2}} + \cdots > \frac{b_{n+1}}{a_{n+1}} \geq \frac{1}{a_{n+1}}$$