Answer. $10 \sqrt{\frac{137}{3}} \approx 67.58$.

Solution. The maximum volume is achieved at the angle $\alpha=60^{\circ}$, then the height of the pyramid is $R \operatorname{tg} \alpha=R \sqrt{3}$, where $R$ is the radius of the circumscribed circle around the triangle. Under the given conditions, there are two possible triangles - an acute-angled and an obtuse-angled one, the areas of which coincide (they are equal to $\frac{5 \cdot 8}{2} \cdot \frac{4}{5}=16$), but the radius of the circumscribed circle will be larger for the obtuse-angled triangle. Then

$$
B C^{2}=5^{2}+8^{2}-2 \cdot 5 \cdot 8 \cdot\left(-\frac{3}{5}\right)=25+64+48=137, \quad \text { and } \quad R=\frac{\sqrt{137}}{2 \cdot \frac{4}{5}}=\frac{5 \sqrt{137}}{8}
$$

The final volume is $V=\frac{1}{3} \cdot 16 \cdot \frac{5 \sqrt{137}}{8} \sqrt{3}=10 \sqrt{\frac{137}{3}} \approx 67.58$.
5.1. What is the maximum volume of the pyramid $S A B C$, where $A B=5, A C=8$ and $\sin \angle B A C = \frac{4}{5}$, and all lateral edges $S A, S B, S C$ form the same angle with the base plane, not exceeding $60^{\circ} ?$