(8) $\frac{\sqrt{6}-1}{10} a$ Hint: Let the radius of the small sphere be $r$, the edge length of the tetrahedron formed by connecting the centers of the four small spheres is $2 r$, and the height is $\frac{2 \sqrt{6} r}{3}$. The height of the regular tetrahedron $S-A B C$ is $\frac{\sqrt{6} a}{3}$, from the tangency relationship we get $4 r+\frac{2 \sqrt{6} r}{3}=\frac{\sqrt{6} a}{3}$, solving for $r$ yields $r=\frac{\sqrt{6}-1}{10} a$.
8 Given a regular tetrahedron $S-ABC$ with edge length $a$, there are four small spheres of the same radius inside it, each touching two others and each touching three adjacent faces, then the radius of the small spheres is $\qquad$