8. If the internal radius $R=2+\sqrt{6} \mathrm{~cm}$ of a hollow sphere contains 4 solid spheres of radius $r$, then the maximum value that $r$ can take is

保留了原文的换行和格式。
Connecting the centers of 4 solid spheres forms a regular tetrahedron, with edge length $2 r$, its height is $\frac{\sqrt{6}}{3} \cdot 2 r=\frac{2 \sqrt{6} r}{3}$, thus the circumradius of the regular tetrahedron $R^{\prime}=\frac{2 \sqrt{6} r}{3} \cdot \frac{3}{4}=\frac{\sqrt{6} r}{2}$. Therefore, $R=2+\sqrt{6}$ $\geqslant R^{\prime}+r=\left(1+\frac{\sqrt{6}}{2}\right) r \Rightarrow r \leqslant 2$. That is, the maximum value of $r$ is 2.