Given the equation about $x$: $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7) \left(\frac{x}{x-1}\right)+1=0$ has real roots.
(1) Find the range of values for $a$;
(2) If the two real roots of the original equation are $x_{1}$ and $x_{2}$, and $\frac{x_{1}}{x_{1}-1} + \frac{x_{2}}{x_{2}-1} = \frac{3}{11}$, find the value of $a$.
Solution: (1) Let $\frac{x}{x-1}=t$, then $t \neq 1$, the original equation becomes $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes $-9 t+1=0$ or $-5 t+1=0$.
That is, $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Then $x=-\frac{1}{8}$ or $x=-\frac{1}{4}$.
Therefore, when $a= \pm 1$, the original equation has real solutions.
When $a \neq \pm 1$, then the original equation has real roots when $\Delta \geqslant 0$.
By $\Delta=[-(2 a+7)]^{2}-4\left(a^{2}-1\right) \geqslant 0$,
Solving gives $a \geqslant-\frac{53}{28}$.
In summary, when $a \geqslant-\frac{53}{28}$, the original equation has real solutions.
The author believes that this solution still misses one case, which is when the original equation has an extraneous root $x=1$, the range of $a$.
Solution: The original equation can be transformed into
$\left(a^{2}-2 a-7\right) x^{2}+(2 a+5) x+1=0$.
When $a^{2}-2 a-7 \neq 0$, in $\Delta \geqslant 0$, the equation has real solutions.
By $\Delta=(2 a+5)^{2}-4\left(a^{2}-2 a-7\right) \geqslant 0$.
Solving gives $a \geqslant-\frac{53}{28}$.
When $a^{2}-2 a-7=0$, i.e., $a=1 \pm 2 \sqrt{2}$, the equation becomes a linear equation, at this time the solution is
$$
x=-\frac{7 \pm 4 \sqrt{2}}{17} \text {. }
$$

Upon verification, $x=-\frac{7 \pm 4 \sqrt{2}}{17}$ meets the requirements and is a solution to the original equation.
Therefore, when $a=1 \pm 2 \sqrt{2}$, the original equation has real roots.
When the original equation produces an extraneous root, i.e., $x=1$, we get $a^{2}-1=0$. Solving gives $a= \pm 1$.
(i) When $a=1$, the original equation solves to $x=-\frac{1}{8}$ or $x=1$. Since $x=1$ is an extraneous root of the original equation, it is discarded. Therefore, $x=-\frac{1}{8}$.
(ii) When $a=-1$, the original equation solves to $x=-\frac{1}{4}$ or $x=1$. Since $x=1$ is an extraneous root of the original equation, it is discarded. Therefore, $x=-\frac{1}{4}$. Hence, when $a= \pm 1$, the original equation has real roots.
In summary, the range of $a$ is $a \geqslant-\frac{53}{28}$.