7. Given $f(x)=8+2 x-x^{2}, g(x)=f\left(2-x^{2}\right)$, then the interval of monotonic increase for the function $g(x)$ is $\qquad$
7. $(-\infty,-1]$ and $[0,1]$
$g(x)=812\left(2-x^{2}\right)-\left(2-x^{2}\right)^{2}=-\left(x^{2}-1\right)^{2}+9$. This function can be seen as the composition of the function $u^{2}+9$ and $u=x^{2}-1$. From $u \leqslant 0$, i.e., $x^{2}-1 \leqslant 0$, we get $-1 \leqslant x \leqslant 1$. The discussion is listed as follows:
\begin{tabular}{|c|c|c|}
\hline$y=-u^{2}+9$ & $u=x^{2}-1$ & $y=-\left(x^{2}-1\right)^{2}+9$ \\
\hline \multirow{2}{*}{$u \in(-\infty, 0]$ is increasing } & Increasing on $[0,1]$ & Increasing on $[0,1]$ \\
\hline & Decreasing on $[-1,0]$ & Decreasing on $[-1,0]$ \\
\hline
\end{tabular}

Continued table
\begin{tabular}{|c|c|c|}
\hline$y=-u^{2}+9$ & $u=x^{2}-1$ & $y=-\left(x^{2}-1\right)^{2}+9$ \\
\hline \multirow{2}{*}{$u \in[0,+\infty)$ is decreasing } & Increasing on $[1,+\infty)$ & Decreasing on $[1,+\infty]$ \\
\cline { 1 - 3 } & Decreasing on $(-\infty,-1)$ & Increasing on $(-\infty,-1]$ \\
\hline
\end{tabular}

Therefore, the monotonic increasing intervals of the function $g(x)$ are $(-\infty,-1]$ and $[0,1]$.