Non Congruent Tilings - 7

This post continues the train of thought in the non-congruent tiling series. The following questions arose from considering how non-congruence of tiles could go over into Elliptic and Hyperbolic geometry (part 6 of the series - http://nandacumar.blogspot.com/2020/04/non-congruent-tilings-6.html). Note: here, we do not always insist on mutual non-congruence of tiles.

Question 1: Is it possible to partition a spherical surface (simply 'sphere' hereafter) into N equal area spherical triangles for any value of N?

Answer: For even N, one can cut a sphere into N identical triangular pieces, the northern hemisphere into N/2 and southern into N/2.
For odd N, there is no vertex to vertex triangulation into N pieces (indeed, for a vertex to vertex triangulation, we have e = 3f/2 => (by Euler formula) v+f = 3f/2 +2. => v = f/2 +2. The last condition cannot be satisfied if f is odd - v has to be an integer).

Question 1a: If the vertex to vertex restriction is relaxed, can we have a triangulation of a sphere into N triangles for any N? Further, can we have triangulations with all pieces having equal area? And equal perimeter? And yes, mutually non-congruent?

Question 2: If we go for spherical quadrilaterals (or pentagons...) instead of triangles, what happens?

Question 3: If we put no restrictions on the number of sides on the pieces: if I understood the experts right, we can have a fair partition of a sphere into N pieces for any N (all N pieces have same area and perimeter) - the proof (by Akopyan, Avvakumov and Karasev) for the existence of a fair partition into N pieces for convex planar regions goes thru for spherical surfaces as well. Can we also achieve piece wise non-congruence?

Question 4: From sphere, one can go over to torus, cube and so on and ask the same questions as above - relaxing the vertex to vertex condition, equal area pieces, equal perimeter pieces, pieces with both area and perimeter equal, then mutually noncongruence added to these conditions.
Note: For a torus, Euler's formula becomes v+f = e => f cannot be odd - because, for an edge to edge triangulation, v = f/2.

Question 5: For what values of N are there N-faced polyhedrons such that all faces have the same area and same perimeter (and perhaps same number of sides, but that may be too strong a constraint) and are mutually non-congruent? Will allowing candidate polyhedrons to be non-convex make the question easier (guess: probably not)? What about relaxing the equal perimeter (or equal area) constraint alone among the faces?

------------
Some Basic Results on question 5:
1. It is easy to see that there is no tetrahedron where the 4 faces are triangles of the same area and perimeter but mutually non-congruent. Indeed, each pair of tetrahedral faces share an edge and that plus the faces having same area and perimeter will demand them to be congruent.

2. With mutually non-congruent faces required to have only equal perimeter (areas free), there is no tetrahedron that achieves it.

Proof: Let such a tetrahedron exist and let a, b, c, d, e, f be the lengths of its 6 edges {a,b,c}, {a,d,e}, {b,e,f} and {c,d,f} being the 4 faces as in this picture.

Since all faces should have equal perimeter, we have a+b+c = a+d+e = b+e+f = c+d+f
Taking these equations pairwise, we have the 6 equations:
b+c = d+e
a+d = b+f
b+e = c+d
a+c = e+f
a+b = d+f
b+e = c+f

Now, a face of the tet can have at most 1 edge length in common with another face (if two triangles with equal perimeter also have two sides each with same lengths, then, the third edge will have the same length - due to the equal perimeter constraint. And if two faces have the sets of edge lengths identical, then the faces are congruent). Faces {a,d,e} and {a,b,c} share a. So, we infer that {d,e} has no value in common with {b,c}. Same is the case with the following pairs of edge pairs:
{e,f} and {a,c}
{d,f} and {a,b}
{a,e} and {c,f}
{b,f} and {a,d}
{d,c} and {b,e}
From this property, we infer that a,b,c,d,e,f have to be all different.

From two of the set of 6 equations we noted earlier, select 2:
b+c = d+e
b+e = c+d
Together these 2 imply that e-c = c-e => c=e, which contradicts the above requirement that a,b,c,d,e,f are all different. Done.

3. Guess: Tetrahedrons exist such that faces are pairwise non-congruent and all have same area.

Justification: Two faces with same area plus two side lengths common will be congruent. So, the 6 relations with edge pairs given above for equal perimeter faces case apply here as well. ie. each of the following pairs of edge pairs have no value in common.
{d,e} and {b,c}
{e,f} and {a,c}
{d,f} and {a,b}
{a,e} and {c,f}
{b,f} and {a,d}
{d,c} and {b,e}
So, all edge lengths: a,b,c,d,e,f are different.

Now, Consider a base edge of the tet, say a. Consider the set of triangles with a as base and some specified area A. The third vertex of this triangle can be any point on an infinitely long cylinder with a as axis. Since a,b,c are different, if we form triangles of same area A with each of a,b,c as base, we have three cylinders with a,b,c as axis and with each cylinder having a unique suitable value of radius. Now, it looks pretty obvious that for some value(s) of the area, the radii of the 3 cylinders will be such that the three cylinders all pass through a common point. That's what we need.

Remark: For number of faces above 4, these questions seem much harder.