8. In the tetrahedron $S-ABC$, $SA=4$, $SB \geqslant 7$, $SC \geqslant 9$, $AB=5$, $BC \leqslant 6$, $AC \leqslant 8$. Then the maximum volume of the tetrahedron is $\qquad$
8. $8 \sqrt{6}$.

Let $\angle S A B=\alpha$.
By the cosine rule, we have
$$
\cos \alpha=\frac{S A^{2}+A B^{2}-S B^{2}}{2 S A \cdot A B} \leqslant-\frac{1}{5} .
$$

Thus, $\sin \alpha=\sqrt{1-\cos ^{2} \alpha} \leqslant \frac{2 \sqrt{6}}{5}$,
$$
S_{\triangle S A B}=\frac{1}{2} S A \cdot A B \sin \alpha \leqslant 4 \sqrt{6} \text {. }
$$

Note that, the height of the pyramid does not exceed its lateral edge length.
Therefore, $V_{\text {pyramid } C-S A B} \leqslant \frac{1}{3} S_{\triangle S A B} \cdot B C \leqslant 8 \sqrt{6}$.
In fact, when $S B=7, B C=6$, and $C B \perp$ plane $S A B$, the given conditions are satisfied, at this time,
$$
V_{\text {pyramid } C-S A B}=8 \sqrt{6} \text {. }
$$