2 As shown in Figure 19, $GF // B_1 D_1, EH // AC // A_1 C_1$, so $GF \perp EH$. Since the projection of $GH$ on plane $ABCD$ is $EH$, it follows that $GH \perp GF$. Similarly, $EF \perp EH$. Therefore, $E$ and $G$ both lie on the sphere with diameter $FH$. Thus, the circumradius of tetrahedron $EFGH$ is $=\frac{1}{2} FH = \frac{\sqrt{2}}{2} a = GF$.
$$
\begin{array}{l}
S_{\triangle FGH} = S_{\triangle FEH} = \frac{1}{2} \times GF \times FH \times \sin 60^\circ = \frac{\sqrt{3}}{4} a^2. \\
S_{\triangle EFG} = S_{\triangle GEH} = \frac{1}{2} \times EG \times GF = \frac{1}{4} a^2.
\end{array}
$$

Therefore, the surface area of tetrahedron $EFGH$ is
$$
S = \frac{\sqrt{2} + \sqrt{3}}{2} a^2.
$$

Let the volume of tetrahedron $EFGH$ be $V$, then
$$
V = \frac{1}{3} \cdot S_{\triangle EFG} \cdot EH = \frac{1}{3} \cdot \frac{\sqrt{2}}{4} a^2 \cdot \frac{\sqrt{2}}{2} a = \frac{1}{12} a^3.
$$

Since the volume of the tetrahedron $V = \frac{1}{3} S r$, we have
$$
\frac{1}{12} a^3 = \frac{1}{3} \cdot \frac{\sqrt{2} + \sqrt{3}}{2} a^2 \cdot r
$$

Thus, $r = \frac{1}{2(\sqrt{2} + \sqrt{3})} a$. Therefore,
$$
\frac{R}{r} = 2 + \sqrt{6}.
$$
$2 E, F, G, H$ are the midpoints of the edges $A B, A_{1} D_{1}$, $A_{1} B_{1}, B C$ of a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$. Try to find the ratio of the radii of the circumscribed sphere and the inscribed sphere of the tetrahedron $E F G H$.