(7) $16 \pi$ Hint: $\left|z_{1}^{4}+1-2 z_{1}^{2}\right|=\left|z_{1}-1\right|^{2}\left|z_{1}+1\right|^{2}$, let $z_{1}=$ $x+y \mathrm{i}$, then we have $x^{2}+y^{2}=1$,
$$
\begin{aligned}
\left|z_{1}^{4}+1-2 z_{1}^{2}\right| & =4(1-x)(1+x) \\
& =4\left(1-x^{2}\right),(-1 \leqslant x \leqslant 1) .
\end{aligned}
$$

It is easy to know $\left|z_{1}^{4}+1-2 z_{1}^{2}\right|_{\text {max }}=4,\left|z_{1}^{4}+1-2 z_{1}^{2}\right|_{\text {min }}=0$, so $0 \leqslant \mid z-7-$ $8 \mathrm{i} \mid \leqslant 4$, thus the region formed by the points corresponding to the complex number $z$ is a circular area with the point $(7,8)$ as the center and 4 as the radius, its area is $16 \pi$.
(7) Let $S=\left\{z|| z-7-8 \mathrm{i}|=| z_{1}^{4}+1-2 z_{1}^{2} \mid ; z, z_{1} \in \mathbf{C}\right.$, $\left.\left|z_{1}\right|=1\right\}$, then the area of the region in the complex plane corresponding to $S$ is