$$
\left.x \in \mathbb{R} \mid \log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right] \in \mathbb{R}\right\}
$$
Now, $\quad(1.6)^{1-x^2}>(0.625)^{6(1+x)}$
$$
\begin{array}{ll}
\Rightarrow & (1.6)^{1-x^2}>(0.625)^{6(1+x)} \\
& =\left(\frac{8}{5}\right)^{1-x^2}>\left(\frac{8}{5}\right)^{-6(1+x)} \\
\therefore & 1-x^2>-6(1+x) \\
\Rightarrow & x^2-6 x-7 < 0 \\
\Rightarrow \quad & (x-7)(x+1) < 0 \\
\Rightarrow \quad & x \in(-\infty,-1) \cup(7, \infty)
\end{array}
$$
Hence,
$$
\begin{gathered}
\left.x \in \mathbb{R}\left|\log \left[(1.6)^{1-x^2}-(0.625)^{6(1+x)}\right]\right| \in \mathbb{R}\right\} \\
=(-\infty,-1) \cup(7, \infty)
\end{gathered}
$$