For a set of 6 edge values, up to 60 distinct tetrahedra can be made in 30 mirrored pairs.
Can a tetrahedron with a given edge set exist entirely inside a second tetrahedron with the same edge set?
Here's a picture of a sample sixty tetrahedra with the same edge set, all with the same centroid.
For this picture $x^3-8 x^2+8=0, x\approx 1.0748$ and the edgeset is $(x^0,x^1,x^2,x^3,x^4,x^5)$. The local minimum for the volume difference between the largest and smallest power-edged tetrahedra seems to be at $12-14 x^2-8 x^4-9 x^6-11 x^{10}-12 x^{12}+13 x^{14}+14 x^{16} =0$, with $x\approx 1.0743129936059068432$
The edgeset (7,8,9,10,11,12) also gives 60 distinct tetrahedra. The edgeset (2,3,4,5,6,7) only yields 8.
Here's a ridiculously hard question.
Can 60 distinct tetrahedra with the same edgeset be solidly packed together to make a convex polyhedron?
If the edge lengths are distinct the longest edge must coincide, so inside must be interpreted as including the surface.
To reduce the number of variables when setting up a search, it's convenient to place one vertex at the origin and one vertex on the $x$-axis. Wlog let this edge along the $x$-axis be the longest and of length $1$; one other vertex be on the $xy$-plane in the positive quadrant; and the final vertex have positive $z$. Then it may be necessary to rotate one of the tetrahedra around the $x$-axis.
I set up a computer search, and on the basis of KISS I didn't initally include the rotation around the $x$-axis. I generate the five remaining lengths randomly, and get a moderate number of hits. Out of 100000 randomly generated edge sets, 1298 give one or more solutions.
For a concrete example, edge lengths $1$, $0.491608192441803$, $0.825600445654988$, $0.472933445811706$, $0.594917355382311$, $0.426403530140595$ give tetrahedra $(0, 0, 0); (1, 0, 0); (0.2800313, 0.4040558, 0); (0.4348697, 0.04946157, 0.1791869)$ and $(0, 0, 0), (1, 0, 0), (0.2800313, 0.4040558, 0), (0.5860534, 0.05409243, 0.08684553)$ which share three vertices but where the fourth vertex of the second tetrahedron is well inside the first tetrahedron.
A solution which only has two vertices in common is $1$, $0.762268094235225$, $0.458549573765392$, $0.695390043172701$, $0.471136764842615$, $0.240386575106711$ with tetrahedron $(0, 0, 0); (1, 0, 0); (0.6853924, 0.3336013, 0); (0.6307988, 0.213046, 0.200678)$ containing tetrahedron $(0, 0, 0); (1, 0, 0); (0.2383665, 0.03109829, 0); (0.6366498, 0.2551774, 0.1145811)$; I haven't calculated how far it can be rotated while remaining inside, but it must be a non-zero amount, so you can lift the third vertex off the $xy$-plane and make the longest edge the only point of surface contact.