9. (16 points) Given that the odd function $f(x)$ is increasing on the interval $(-\infty, 0)$, and $f(-2)=-1, f(1)=0$, when $x_{1}>0, x_{2}>0$, we have $f\left(x_{1} x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)$, find the solution set of the inequality $\log _{2}|f(x)+1|<0$.
9. From $\log _{2}|f(x)+1|<0$, we get $0<|f(x)+1|<1$, so $-1<f(x)+1<0$ or $0<f(x)+1<1$, which means $-2<f(x)<-1$ or $-1<f(x)<0$.

Given that $f(x)$ is an odd function and is increasing on the interval $(-\infty, 0)$, we know that $f(x)$ is increasing on $(0,+\infty)$, and $f(2)=1$. Thus, $f(1)=f\left(2 \cdot \frac{1}{2}\right)=f(2)+f\left(\frac{1}{2}\right)$, so $f\left(\frac{1}{2}\right)=-1, f\left(\frac{1}{4}\right)=$ $f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=-2$.

Also, $f(4)=f(2)+f(2)=2$, and since $f(x)$ is an odd function, $f(-4)=-f(4)=-2, f(-2)=-f(2)=$ $-1, f(-1)=-f(1)=0$.

Therefore, $x$ should satisfy $-4<x<-2$ or $-2<x<-1$ or $\frac{1}{4}<x<\frac{1}{2}$ or $\frac{1}{2}<x<1$, which means the solution set is $(-4,-2) \cup(-2,-1) \cup\left(\frac{1}{4}, \frac{1}{2}\right) \cup\left(\frac{1}{2}, 1\right)$.