Example 12 Let $f(x)$ be an odd function defined on $(-\infty, 0) \cup(0,+\infty)$, and it is increasing on $(0,+\infty)$.
(1) Determine the monotonicity of $f(x)$ on $(-\infty, 0)$, and prove it using the definition of a monotonic function;
(2) If $f(1)=0$, solve the inequality $\int\left[\log \left(1-x^{2}\right)+1\right]>0$ for $x$, where $a>1$;
(3) If $m>0, n>0$ and $f(m \cdot n)=f(m)+f(n)$. Given $f(-2)=-1$, find the range of $t$ for which $\log _{4}|f(t)+1|>$ () holds.
$$
\begin{array}{c}
\text { (1) Let } x_{1}-x_{2}>0. \\
\text { Since } f(x) \text { is an increasing function on } (0,+\infty), \\
\text { then } f\left(-x_{1}\right)>f\left(-x_{0}\right). \\
\text { Also, since } f(x) \text { is an odd function, } -f\left(x_{1}\right)>-f\left(x_{2}\right). \\
\text { Hence } f\left(x_{\mathrm{t}}\right)0, \\
\log _{u}\left(1-x^{2}\right)+1>1,
\end{array}\right. \\
\left\{\begin{array}{l}
\log _{a}\left(1-x^{2}\right)+1-1 .
\end{array}\right.
\end{array}\right.
$$

or
$$
\begin{array}{c}
\left\{\begin{array}{l}
\log _{u}\left(1-x^{2}\right)+1>0, \\
\log _{u}\left(1-x^{2}\right)+1>1,
\end{array}\right. \\
\left\{\begin{array}{l}
\log _{4}\left(1-x^{2}\right)+1-1 .
\end{array}\right.
\end{array}
$$
(1) The equation is $\log _{\alpha}\left(1-x^{2}\right)>0$.

Since $a>1$, then $1-x^{2}>1$ which means $x^{2}1$,
so $\frac{1}{a^{2}}0$,
so $0<1 f(t)+1 \mid<1$.
So $-2<f(t)<-1$ or $-1<f(t)<0$.
Based on the monotonicity of the function $f(x)$, we only need to find the $t$ values corresponding to $-2,-1,0$ to determine the range of $t$.
Since $f(1)=f(1)+f(1)$,
then $f(1)=0$ and $f(-1)=-f(1)=0$.
Also, since $f(-2)=-1$,
then $f(2)=1, f(4)=f(2)+f(2)=2, f(-4)=-f(4)=-2$.
Also, since $f(2)=f(4)+f\left(\frac{1}{2}\right)$.
Then $f\left(\frac{1}{2}\right)=-1, f\left(\frac{1}{4}\right)=f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=-2$.
Therefore, the range of $t$ is
$$
(-4,-2) \cup(-2,-1) \cup\left(\frac{1}{4}, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 1\right) \text {. }
$$
$$