15. Given the equation about $x$

Figure 3 $\left(a^{2}-1\right)\left(\frac{x}{x-1}\right)^{2}-(2 a+7)\left(\frac{x}{x-1}\right)+1=0$ 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 $a$.
15.1) Let $\frac{x}{x-1}=t$, then $t \neq 1$. The original equation can be transformed into $\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0$.
When $a^{2}-1=0$, i.e., $a= \pm 1$, the equation becomes
$-9 t+1=0$ or $-5 t+1=0$,
which means $\frac{x}{x-1}=\frac{1}{9}$ or $\frac{x}{x-1}=\frac{1}{5}$.
Thus, $x=-\frac{1}{8}$ or $x=\frac{1}{4}$.
Therefore, when $a= \pm 1$, the original equation has real roots.
When $a \neq \pm 1$, the original equation has real roots if $\Delta \geqslant 0$.
By $\Delta=[-(2 a+7)]^{2}-4\left(a^{2}-1\right) \geqslant 0$,
we get $a \geqslant-\frac{53}{28}$.
In summary, when $a \geqslant-\frac{53}{28}$, the original equation has real solutions.
(2) From the given, $\frac{x_{1}}{x_{1}-1}$ and $\frac{x_{2}}{x_{2}-1}$ are the two roots of the equation
$$
\left(a^{2}-1\right) t^{2}-(2 a+7) t+1=0
$$

Using the relationship between roots and coefficients, we get $\frac{2 a+7}{a^{2}-1}=\frac{3}{11}$. Thus,
$$
3 a^{2}-22 a-80=0,
$$

Solving this, we get $a_{1}=10, a_{2}=-\frac{8}{3}$.
From (1), we know that $a \geqslant-\frac{53}{28}$.
Since $a_{2}=-\frac{8}{3}<-\frac{53}{28}$, $a_{2}$ should be discarded.
Therefore, $a=10$.