3. Given the function $f(x)=4 \pi \arcsin x-(\arccos (-x))^{2}$ with the maximum value $M$ and the minimum value $m$. Then $M-m=$ $\qquad$ .
3. $3 \pi^{2}$.

Since $\arccos (-x)=\pi-\arccos x, \arcsin x=\frac{\pi}{2}-\arccos x$, we have
$$
\begin{aligned}
f(x) & =4 \pi\left(\frac{\pi}{2}-\arccos x\right)-(\pi-\arccos x)^{2} \\
& =-(\arccos x)^{2}-2 \pi \arccos x+\pi^{2} \\
& =-(\arccos x+\pi)^{2}+2 \pi^{2} .
\end{aligned}
$$

Given $\arccos x \in[0, \pi]$, we know that the maximum value of $f(x)$ is $\pi^{2}$, and the minimum value is $-2 \pi^{2}$. Therefore, $M-m=3 \pi^{2}$.