37. The faces of a parallelepiped with edges of length $a$ are rhombuses with an acute angle of $60^{\circ}$. Calculate the volume of the sphere inscribed in the parallelepiped.
37. Given that the magnitude of the acute angles of all faces is $60^{\circ}$, the diagonals of the faces are congruent to the lateral edges of the parallelepiped. Let $A B C D$ be one of the faces, $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ - the parallel face, and $A$ - the vertex of three acute angles. If two regular tetrahedra $A A^{\prime} B D$ and $C C^{\prime} B^{\prime} D^{\prime}$ are separated from this parallelepiped, an octahedron $A^{\prime} B D C B^{\prime} D^{\prime}$ will remain.

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Fig. 116 (prove). The sphere inscribed in this octahedron will touch all the faces of the parallelepiped. From problem 31, we have that if $d$ is the length of half the diagonal of the octahedron, then the radius $r$ of the inscribed sphere is determined by the formula: $d=r \sqrt{3}$. The length of the edge of the octahedron is $a=d \sqrt{2}$. Therefore, $a=r \sqrt{6} ; r=\frac{a}{\sqrt{6}}$, and the volume of the sphere $V=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \frac{a^{3}}{6 \sqrt{6}}=\frac{2}{9} \pi \frac{a^{3}}{\sqrt{6}}$.