Solution:

(a) Let $P_{1}$ be the number of those who passed before the increase and $P_{2}$ the number of those who passed after the increase in score. From the information we have we can write:
$$66 N = 71 P_{1} + 56 (N - P_{1}), \quad 71 N = 75 P_{2} + 59 (N - P_{2})$$
From the first relation, carrying out the computations, we obtain $10 N = 15 P_{1}$ or $2 N = 3 P_{1}$, from which, since $N$ and $P_{1}$ are integers, we can conclude that $N$ is a multiple of 3. Similarly, from the second relation we have $12 N = 16 P_{2}$ or $3 N = 4 P_{2}$, and hence $N$ is a multiple of 4. In conclusion, $N$ must be a multiple of 3 and of 4, that is of 12, and can therefore be 12, 24 or 36. These three cases are indeed possible. Let us look for an example for $N = 12$, trying to have grades as equal as possible. If 8 people, that is those who passed right from the start, have a grade before the increase of 71, one has taken 62 and the remaining three 54, we have verified all the hypotheses of the problem. The cases $N = 24$ and $N = 36$ are analogous, respectively doubling and tripling the people in each score band.

(b) No $N$ can satisfy the hypotheses of this point. Indeed, let $M_{p}$ be the average - before the increase - of those who were NOT passing before the increase but who would become so after. Then we would have:
$$76 P_{1} + (M_{p} + 5)(P_{2} - P_{1}) = 79 P_{2}$$
But $M_{p} < 65$, hence $M_{p} + 5 < 70 < 76$, from which
$$79 P_{2} = 76 P_{1} + (M_{p} + 5)(P_{2} - P_{1}) < 76 P_{1} + 76 (P_{2} - P_{1}) = 76 P_{2} < 79 P_{2}$$
absurd.