For what values of the parameter $a$ does the equation $\frac{\log _{a} x}{\log _{a} 2}+\frac{\log _{x}(2 a-x)}{\log _{x} 2}=\frac{1}{\log _{\left(a^{2}-1\right)} 2}$ have:
(1) solutions?
(2) exactly one solution?
The original equation is equivalent to
$$
\left\{\begin{array} { l } 
{ a ^ { 2 } - 1 > 0 } \\
{ a ^ { 2 } - 1 \neq 1 } \\
{ x \neq 1 } \\
{ x > 0 } \\
{ \operatorname { log } _ { 2 } x + \operatorname { log } _ { 2 } ( 2 a - x ) = \operatorname { log } _ { 2 } ( a ^ { 2 } - 1 ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x>0 \\
x \neq 1 \\
2 a-x>0 \\
x(2 a-x)=a^{2}-1
\end{array}\right.\right.
$$

The equation $x^{2}-2 a x+a^{2}-1=0$ has solutions $x_{1}=a+1, x_{2}=a-1$. Therefore,
（1）If $x$ has a solution, then $0 < a-1 < 2a$ and $0 < a+1 < 2a$, and $a > 1$, and $a \neq \sqrt{2}$. Thus, the original equation has a solution when $a > 1$ and $a \neq \sqrt{2}$.
（2）If there is only one solution, then
$$
\left\{\begin{array} { l } 
{ a + 1 \in ( 0 , 2 a ) } \\
{ a - 1 \notin ( 0 , 2 a ) }
\end{array} \text { or } \left\{\begin{array} { l } 
{ a + 1 \in ( 0 , 2 a ) } \\
{ a - 1 = 1 }
\end{array} ; \text { or } \left\{\begin{array} { l } 
{ a + 1 \notin ( 0 , 2 a ) } \\
{ a - 1 \in ( 0 , 2 a ) }
\end{array} \text { or } \left\{\begin{array}{l}
a+1=1 \\
a-1 \in(0,2 a)
\end{array}\right.\right.\right.\right.
$$

Thus, the solution is $a=2$.