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Provided $p$ divides all of those binomials, there's no obvious obstruction. So they seem possible, but they have not been constructed or shown to exist. And by the same criteria, a projective plane of order 10 seems possible, and isn't. Your best bet for finding a counterexample is the case $k=3$, and your best bet for proving they all exist is time travel.
There's a slightly larger example using $p=5$, $n=8$ and $k=4$, where each element is in $7$ sets, each pair is in $3$, and each triple is in $1$. It basically designs itself: take any set of size $4$ (e.g. $1234$), the complement of that (so $5678$), and then two sets for each pair in the first set ($12xx$, $12xx$), filled in using disjoint parts of the complement (like $1256$, $1278$) so that all three partitions of the complement are used for each element in the first set.
It sounds like you're asking for the best algorithm to reverse a queue using two other FIFO queues. It might be better to ask on stack exchange - a cursory glance at a "reverse queue" search there makes it look popular.
The $2^4$ hypercube is impossible: the corresponding graph has two maximum independent sets of size 8 (via 2-colourings), so even numbers must go there. Then any two vertices have at most two common neighbours, so there are at most two spots not adjacent to either 6 or 12.
No, it isn't. Take $u_3$ to be an odd term in oeis.org/A000129, the denominators of the best (lower) rational approximations to $\sqrt{2}$. Then $u_1$ and $u_2$ differ by $1$, the terms grow exponentially, and you can generate as many as you want. Pell equations, and continued fractions, are the way to think about these.
Your question is "what is the maximal $k$ which makes SGP ($10-7-k$) solvable". The SGP does not ask for a perfect solution, as the author notes on page 7 of the freely available thesis, and which should be clear from the simpler 10 week solution for 8 groups of 4. The thesis also includes an algorithm in ECLiPSe (page 48) for arbitrary numbers of golfers and groups, an SAT implementation, and gives a link to Warwick Harvey's page listing known (as of 2002) solutions, including 5 compatible partitions for yours. It really deserves more than a glance if you are researching this topic.
My comment above isn't accurate in this case, since there will be pairs of people who are never in the same group, so it isn't a complete block design. I should also add that the thesis results have been published in Ann. Operations Research (2012), link.springer.com/article/10.1007%2Fs10479-011-0866-7
This is an example of a "social golfer problem", which asks for a maximal (70,10,1) block design with parallelism. You can read up on solution methods in this master's thesis, logic.at/prolog/sgp/sgp.html
Equating coefficients (in either the case of $4$ or $k$) gives you two sums of squares equations, and one sum of products. Compare these with the Cauchy-Schwarz inequality to confirm that they must all be scalar multiples.
Your main question is the 4 dimensional version, colouring the 8 cubes of tesseracts, and stacking them into a $4\times4\times4\times4$ tesseract. I don't think it does much directly to rephrase it this way, but you might find some restrictions in the 3 dimensional case that carry over.
Do you have a solution to the intermediate problem coming from $2^6 = 4^3$? i.e. paint the sides of cubes with two colours, then arrange the $64$ distinct cubes into one large cube with all outer faces the same colour.
