13. In a regular tetrahedron $S-ABC$, the dihedral angle between two adjacent lateral faces is $2 \alpha$, and the distance from the center of the base $O$ to a lateral edge is 1. Find $V_{\text {S }}$.
13. $S O$ is the perpendicular line to plane $A B C$, and let plane $S O A$ intersect $B C$ at $D$.

In plane $S A D$, construct $O E \perp S A$, with $E$ as the foot of the perpendicular, then $O E=1$.
Construct a plane perpendicular to $S A$ through $O E$, intersecting $A B, A C$ at $B_{1}, C_{1}$, respectively. Then $B_{1} E, C_{1} E$ are both perpendicular to $S A$,
so $\angle B_{1} E C_{1}$ is the plane angle of the dihedral angle $B-S A-C$.
By symmetry, it is easy to see that $\angle B_{1} E O=\angle O E C_{1}=\alpha, E B_{1}=E C_{1}, A B_{1}=A C_{1}$,
so $B_{1} C_{1} / / B C$. The intersection line $B_{1} C_{1}$ of plane $E B_{1} C_{1}$ with $A B C$ passes through $O$, so
$$
B_{1} C_{1}=\frac{2}{3} B C .
$$

In the isosceles triangle $E B_{1} C_{1}$, it is easy to know that $B_{1} C_{1}=2 \tan \alpha$.

For the equilateral triangle $A B C$, the side length, height, and area are respectively
$$
B C=3 \tan \alpha, A D=\frac{3 \sqrt{3}}{2} \tan \alpha, S_{\triangle A B C}=\frac{\sqrt{3}}{4} \cdot(3 \tan \alpha)^{2},
$$

Thus, $A O=\sqrt{3} \tan \alpha$
From $\triangle S A O \sim \triangle O A E$, we have $S O=\frac{O E \cdot A O}{A E}=\frac{1 \cdot \sqrt{3} \tan \alpha}{\sqrt{(\sqrt{3} \tan \alpha)^{2}-1^{2}}}=\frac{\sqrt{3} \tan \alpha}{\sqrt{3 \tan ^{2} \alpha-1}}$.
Thus, $V_{S A B C}=\frac{1}{3} \cdot S O \cdot S_{\triangle A B C}=\frac{9 \tan ^{3} \alpha}{4 \sqrt{3 \tan ^{2} \alpha-1}}$.