Solution:
If $a \geq 2$, then the roots $x_{1}$ and $x_{2}$ of the equation $x^{2}-a x+1=0$ are positive and $x_{1} x_{2}=1$. In particular, $S_{n}>0$ for $n=1,2, \ldots$

a) We have
$$\begin{aligned} \frac{S_{n-1}}{S_{n}} \geq \frac{S_{n}}{S_{n+1}} &\Longleftrightarrow \left(x_{1}^{n-1}+x_{2}^{n-1}\right)\left(x_{1}^{n+1}+x_{2}^{n+1}\right) \geq \left(x_{1}^{n}+x_{2}^{n}\right)^{2} \\ &\Longleftrightarrow x_{1}^{n-1} x_{2}^{n+1}+x_{2}^{n-1} x_{1}^{n+1} \geq 2 x_{1}^{n} x_{2}^{n} \\ &\Longleftrightarrow \left(x_{1} x_{2}\right)^{n-1}\left(x_{1}-x_{2}\right)^{2} \geq 0 \end{aligned}$$
which obviously holds.

b) Let $a \geq 2$ have the desired property. Then a) implies that
$$n \frac{S_{1}}{S_{2}} \geq \frac{S_{1}}{S_{2}}+\cdots+\frac{S_{n}}{S_{n+1}}>n-1$$
i.e., $\frac{S_{1}}{S_{2}}>1-\frac{1}{n}$. Since $\lim _{n \rightarrow \infty} \frac{1}{n}=0$, the last inequality gives $\frac{S_{1}}{S_{2}} \geq 1$. Using Vieta's formulas we get $S_{1}=a$, $S_{2}=a^{2}-2$ and therefore $\frac{a}{a^{2}-2} \geq 1 \Longleftrightarrow \frac{(a+1)(a-2)}{a^{2}-2} \leq 0$. Since $a \geq 2$ we get $a=2$.

Conversely, if $a=2$, then $x_{1}=x_{2}=1$ and $S_{n}=2$ for any $n=1,2, \ldots$ Hence
$$\frac{S_{1}}{S_{2}}+\frac{S_{2}}{S_{3}}+\cdots+\frac{S_{n}}{S_{n+1}}=n>n-1$$