5. (Dongtai City) In the tetrahedron $S-ABC$, $SA$ is perpendicular to the base $ABC$, and the lateral faces $SAB$ and $SBC$ form a right dihedral angle. If $\angle BSC = 45^{\circ}, SB = a$, find the volume of the circumscribed sphere of this tetrahedron.
Let the plane of $\triangle A B C$ be $\gamma$, the plane of $\angle S A B$ be $\alpha$, and the plane of $\triangle S B C$ be $\rho$.
$\because S A \perp \gamma, \therefore S A \perp B C$.
Also, $\alpha \perp P$, draw $A D \perp S B$ in $\alpha$ through $A$, then $A D \perp B$, so $A D \perp B C$, and $S A \cap D A=\Lambda$, which implies $C B \perp S B$.
Therefore, $\triangle S B C$ is a right-angled triangle.
Since $S A \perp \gamma$, then $S A \perp A C, A \_S A C$ is a right-angled triangle.
Take the midpoint $O$ of $S C$, connect $A O, B O$, by the theorem, we have $S O=O C=O A=O B$, so $O$ is the center of the circumscribed sphere of $S-A B C$, with a radius of $\frac{\sqrt{2}}{2} a$. Therefore,
$$
V_{\text {sphere }}=\frac{4}{3} \pi\left(\frac{\sqrt{2}}{2} a\right)^{3}=\frac{\sqrt{2}}{3} \pi a^{3} \text { . }
$$