II. (30 points) Given the function
$$
f(x)=\log _{a} \frac{x-3}{x+3}(a>0 \text {, and } a \neq 1) \text {. }
$$

If there exist real numbers $m, n (m<n)$ and $a$, such that the domain of $f(x)$ is $(m, n)$, and the range is
$$
\left(1+\log _{a}(n-1), 1+\log _{a}(m-1)\right),
$$

find the ranges of $m$ and $a$ respectively.
From $\frac{x-3}{x+3}>0$, we get the range of $x$ as
$(-\infty,-3) \cup(3,+\infty)$.
Since the domain of $f(x)$ is $(m, n)$, and $m>1$, $n>1$, it follows that $m>3$.
Also, $m-13$.
The range of $f(x)$ is
$$
\left(1+\log _{a}(n-1), 1+\log _{a}(m-1)\right) \text {, }
$$

Then $\log _{a} a(n-1)0, \\
g(3)=12 a>0, \\
-\frac{2 a-1}{2 a}>3 .
\end{array}\right.
$$

Solving this, we get $0<a<\frac{2-\sqrt{3}}{4}$.
Therefore, the range of $m$ is $(3,+\infty)$, and the range of $a$ is $\left(0, \frac{2-\sqrt{3}}{4}\right)$.