8. The maximum volume of a cone inscribed in a sphere with radius 1 is . $\qquad$
8. $\frac{32 \pi}{81}$.

Let the radius of the base of the inscribed cone be $r$, and the height be $h$.
To find the maximum volume, by symmetry, it is clear that we should take $h \geqslant 1$, then $h=1+\sqrt{1-r^{2}}$.
Thus, $V=\frac{\pi}{3} r^{2} h=\frac{\pi}{3} r^{2}\left(1+\sqrt{1-r^{2}}\right)$.
Let $d=\sqrt{1-r^{2}} \geqslant 0$. Then
$$
\begin{array}{l}
V=\frac{\pi}{3} r^{2} h=\frac{\pi}{3}\left(1-d^{2}\right)(1+d) \\
=\frac{\pi}{6}(1+d)(1+d)(2-2 d) \\
\leqslant \frac{\pi}{6}\left[\frac{(1+d)+(1+d)+(2-2 d)}{3}\right]^{3} \\
=\frac{32 \pi}{81} .
\end{array}
$$

The equality holds if and only if $1+d=2-2 d \Rightarrow d=\frac{1}{3}, r=\frac{2 \sqrt{2}}{3}$.