a. This is wrong. For example, one might place the numbers from $1$ to $8$ along the main diagonal and the numbers from $57$ to $64$ along the secondary diagonal:

| 1 | 9 | 10 | 11 | 12 | 13 | 14 | 57 |
|----|----|----|----|----|----|----|----|
| 15 | 2 | 16 | 17 | 18 | 19 | 58 | 20 |
| 21 | 22 | 3 | 23 | 24 | 59 | 25 | 26 |
| 27 | 28 | 29 | 4 | 60 | 30 | 31 | 32 |
| 33 | 34 | 35 | 61 | 5 | 36 | 37 | 38 |
| 39 | 40 | 62 | 41 | 42 | 6 | 43 | 44 |
| 45 | 63 | 46 | 47 | 48 | 49 | 7 | 50 |
| 64 | 51 | 52 | 53 | 54 | 55 | 56 | 8 |

Therefore the numbers from $1$ to $8$ are column minima, whereas the numbers from $57$ to $64$ are row maxima. Therefore, no number is at the same time column minimum and row maximum, so no number is super-plus-good.

b. This is true. Denote the number in the $a$th row and $b$th column by $F(a, b)$. Assume that there exist two super-plus-good numbers, and let $(i, j)$ and $(r, s)$ be the coordinates of these two numbers. Since all numbers are different, the row maxima and column minima are unique. Therefore no row and no column may contain more than one super-plus-good number, so $i \neq r$ and $j \neq s$ must hold. Then
$$F(i, j) > F(i, s) \quad (\text{because } F(i, j) \text{ is row maximum}).$$
$$F(i, j) < F(r, j) \quad (\text{because } F(i, j) \text{ is column minimum}).$$
$$F(r, s) > F(r, j) \quad (\text{because } F(r, s) \text{ is row maximum}).$$
$$F(r, s) < F(i, s) \quad (\text{because } F(r, s) \text{ is column minimum}).$$
These four inequalities lead to the following contradiction:
$$F(i, j) > F(i, s) > F(r, s) > F(r, j) > F(i, j).$$