There are 15 possible pairs of questions which we can number:
(1) AB (2) BC (3) CD (4) DE (5) EF
(6) AC (7) BD (8) CE (9) DF
(10) AD (11) BE (12) CF
(13) AE (14) BF
(15) AF
We have 36 students, so if we allocate each pair of questions in order to the first 15 students, each question would be allocated to 5 students, and no two students would be given the same pair of questions.
But there are 21 students remaining, so if we allocate question pairs again in order to the next 15 students, we end up with each question being allocated to 10 students.
That leaves 6 students to be allocated a pair of questions each. Out of the 15 combinations, the number of ways of selecting 6 out of 15 is 5005.
The end result is that each question would be allocated to at least 10 students.
Looking at the requirements of the problem, the last requirement needs clarification. It’s nevertheless difficult to see how there can be a solution to the problem as it has been presented.