Rax $* \star$ Given that the wooden tetrahedron has three pairs of opposite edges all equal, each being $13 \mathrm{~cm}$, $14 \mathrm{~cm}$, and $15 \mathrm{~cm}$. If it is to be carved into a sphere, what is the maximum radius of the sphere?
As shown in the figure, without loss of generality, let $A C=B D=13 \mathrm{~cm}$, $B C=A D=14 \mathrm{~cm}$, $A B=C D=15 \mathrm{~cm}$. Let $E, F, G, H, K, L$ be the midpoints of the edges of the tetrahedron. Then $E F=G H=\frac{1}{2} A C=\frac{1}{2} B D=E H=F G$, and $E F \parallel A C \parallel H G$. Therefore, $E F G H$ is a rhombus, and its diagonals $E G$ and $F H$ are perpendicular and bisect each other at point $O$. Similarly, the midpoints of the three pairs of opposite edges all intersect perpendicularly and bisect at point $O$, thus forming four congruent tetrahedra $O-E F K, O-F G L, O-G H K, O-H E L$. The heights of these tetrahedra are the radius $r$ of the inscribed sphere (the largest sphere) of the tetrahedron.
From the fact that the diagonals of a rhombus are perpendicular and bisect each other, we have
$$
\left\{\begin{array}{l}
O G^{2}+O F^{2}=G F^{2}-\left(\frac{13}{2}\right)^{2}, \\
O G^{2}+O L^{2}=G L^{2}=\left(\frac{14}{2}\right)^{2}, \\
O L^{2}+O F^{2}=L F^{2}=\left(\frac{15}{2}\right)^{2}
\end{array}\right.
$$

Solving these equations, we get $O G=\frac{\sqrt{70}}{2}, O F=\frac{3 \sqrt{11}}{2}, O L=\frac{3 \sqrt{14}}{2}$. Thus, the volume of the tetrahedron $O-F G L$ is $V=\frac{1}{3} \times \frac{1}{2} \times \frac{3 \sqrt{11}}{2} \times \frac{3 \sqrt{14}}{2} \times \frac{\sqrt{70}}{2}=\frac{21 \sqrt{55}}{8}$. The area of the base $\triangle F G L$ is 21, so $\frac{1}{3} r \cdot 21=\frac{21 \sqrt{55}}{8}, r=\frac{3 \sqrt{55}}{8} \mathrm{~cm}$.