For the ones you said "???", i would suggest something like this (but im not a maths student so i dont know how much detail is supposed to be in the answer):
(2) i dont think you can deduce the number of rows in A. Proof by contradiction: Suppose you knew A and it had m rows, if you added a copy of the last row the same vector x would still solve Ax=d. Alternatively (and trivially) add a row of zeros to both A and d. All that can be deduced is the number of linearly independant rows in A (which is the next question)
There is only one solution iff there are 5 unique rows (Rouché–Capelli theoum according to wikipedia). Your system has infinitely many solutions (corresponding to all possible values of s and t). each "spare" variable means the rank of A must be reduced by 1.
This could be proved by adding 2 new (linearly independant) equations for s,t and noting that there is then only 1 solution.
I'll leave it to someone better than me to verify your other answers.