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The short answer is "no, not really". In general, there are virtually no positive 3- (and higher) dimensional results on tilings, see my old survey, section 8 on a few sporadic results. Instead, mos …

You are perhaps thinking of Tutte's "spring" theorem that every 3-connected graph has an embedding with convex faces. It indeed a corollary of the (earlier) Steinitz theorem, but Tutte's proof is of …

There are only partial answers to this question. First, one can prove that Walkup's result cannot be proved using coloring arguments (I think I did this in New horizons paper, but the setting is form …

Now take $nQ = [nst \times nst]$ and a standard "brick tiling" of $nQ$, with $stn^2$ copies of translates of $[s\times t]$. There are $\theta(n)$ lines to be "blocked". … For every copy of $Q$ inside $nQ$, we can "flip" from one tiling to another. This will block some constantly many lines. …

Curiously, any polytope which tiles $\Bbb R^3$ or $\Bbb R^4$ must be scissors congruent to a cube, even if the tiling is aperiodic (see here). …

$1)$ To answer your finiteness question - this is pretty standard. There is a finite number of combinatorial arrangements of tiles. Each leads to a system of linear equations which can be solved. …

You don't need an anchor tile. Jed Yang and I recently solved this on the way to stronger NP-completeness results (versions of this result were already known). We explain it all here. Similarly, th …

You might want to consider reading the following excellent introduction to the subject: Charles Radin, "Miles of tiles", AMS, 1999.

Definitely not tetrahedra, see this paper. The proof generalizes to other families of polytopes, actually. For example, by Sydler's theorem these cannot be polytopes with nonzero Dehn invariant ($\L …