(a) For a face $\mathcal F$ of our polyhedron, let $\ell(\mathcal F)$ be the perpendicularto the plane of $\mathcal F$ passing through the center of $\mathcal F$. In order to prove that the polyhedron can be inscribed in a sphere, it suffices to prove that for every vertex $A$ of the polyhedron, all $\ell(\mathcal F)$ as $\mathcal F$ ranges through the faces containing $A$ intersect. The angles of a face with $n$ sides are $\frac{n-2}n\pi\ge\frac\pi 2$, and the angles at $A$ of the faces containing $A$ must sum up to less than $2\pi$. This means that we can never have at least four faces meeting in $A$, or, in other words, for every vertex $A$ of the polyhedron, exactly three faces meet in $A$. If the three edges passing through $A$ are $AA_{1},AA_{2},AA_{3}$ and the faces containing $A$ are $\mathcal F_{1},\mathcal F_{2},\mathcal F_{3}$, contained in the planes $(AA_{2}A_{3}),(AA_{3}A_{1}),(AA_{1}A_{2})$ respectively, then $\ell_{\mathcal F_{i}}$ intersect simply because a tetrahedron (in this case $AA_{1}A_{2}A_{3}$) can be inscribed in a sphere. (b) Start placing the faces of our polyhedron as follows: place the first one arbitrarily, then one which is adjacent to the first one (“adjacent” meaning that they share a side), and then, from the third step onwards, each new face must have a vertex in common with two faces which have already been placed. Suppose in the current step we are placing down the face $\mathcal F$, adjacent to two previously posted faces $\mathcal F_{1},\mathcal F_{2}$ (which share a side). If $\mathcal F_{1},\mathcal F_{2}$ are the first two faces, then $\mathcal F_{1},\mathcal F_{2},\mathcal F$ are of at most three types. If, on the other hand, $\mathcal F_{1},\mathcal F_{2}$ are not the first two faces, then the other face which is adjacent to both $\mathcal F_{1},\mathcal F_{2}$ has already been placed, and, by symmetry with respect to the perpendicular bisector plane of the common side of $\mathcal F_{1},\mathcal F_{2}$, it must be congruent to $\mathcal F$. This shows that from the fourth step onwards all faces which appear must be of a type which has appeared before, and proves the assertion in (b). (c) It suffices to prove that in some point of the polyhedron two hexagons and one pentagon meet, because then, by using a procedure of placing down the faces one by one as described above, there is only one polyhedron we can obtain (in other words, a choombam is uniquely determined by the three faces which meet at any one of its vertices). The three faces meeting at a point cannot all be hexagons, because then the angles would add up to $2\pi$, and they must add up to less than $2\pi$. If they are all pentagons, then the polyhedron is the regular dodecahedron. It is also impossible that they be two pentagons and a hexagon. In fact, in general, the neighboring faces of a face with an odd number of sides in a choombam must all be congruent, as can easily be seen. This leaves just one possibility: the three faces meeting at a vertex are two hexagons and a pentagon, and this is precisely what we wanted to prove. (d) Again, we make use of the observation made above that a choombam is uniquely determined by the triple of plane angles meeting at any one of its points. This means that there is an injection from the set of choombams into the set of triples of the form $\left(\frac{n-2}n,\frac{m-2}m,\frac{p-2}p\right)$ with $m,n,p\ge 4$ and such that $\frac{n-2}n+\frac{m-2}m+\frac{p-2}p<2$. If such a triple contains two angles $\frac12$, then the corresponding choombam is a prism, as described in Omid’s post. Since there are only finitely many such triples with only one $\frac{1}{2}$, we’re done.