10. (20 points) Let the function
$$
f(x)=\log _{a}(1+a x)-\log _{a}(1-a x),
$$

where $a>0$, and $a \neq 1$.
(1) When $a>1$, if the inequality
$$
f(x) \geqslant \log _{a} 8 x
$$

holds for all $x$, find the range of values for $a$;
(2) If $f\left(x_{0}\right)=x_{0}-1$, prove: $\left|x_{0}\right|<1$.
$$
\begin{array}{l}
(1) \text{ Let } a x=t. \text{ Then } 01) \\
\Leftrightarrow \frac{8 t}{a} \leqslant \frac{1+t}{1-t} \\
\Leftrightarrow \frac{a}{8} \geqslant \frac{t(1-t)}{1+t}=-(t+1)-\frac{2}{t+1}+3 .
\end{array}
$$

Thus, $a \geqslant 24-16 \sqrt{2}$.
(2) From $\left\{\begin{array}{l}1-a x>0, \\ 1+a x>0, \\ a>0\end{array} \Rightarrow-\frac{1}{a}1$, then $\left|x_{0}\right|0, \\
F(1)=\log _{a} \frac{1+a}{1-a}<\log _{a} 1=0,
\end{array}
$$

Then by the Intermediate Value Theorem, $x_{0} \in(0,1)$.
Therefore, $\left|x_{0}\right|<1$.
In conclusion, $\left|x_{0}\right|<1$.
$$