8. Let $f(x)=|x+1|+|x|-|x-2|$. Then $f(f(x))+1=0$ has different solutions.

The number of different solutions is:
8.

Notice,
$$
f(x)=\left\{\begin{array}{ll}
-x-3, & x \leqslant-1 ; \\
x-1, & -12 .
\end{array}\right.
$$

From $f(f(x))+1=0$
$\Rightarrow f(x)=-2$ or 0
$\Rightarrow x=-1$ or -3 or $\frac{1}{3}$, a total of three solutions.