A 7 × 7 board has its rows numbered 0 to 6 from the top and its columns numbered 0 to 6 from the…

(a) Among $0, 1, \dots, 6$, three numbers ($0, 3, 6$) leave remainder $0$, two ($1, 4$) leave $1$ and two ($2, 5$) leave $2$. The sum $i + j$ is a multiple of $3$ when the remainders of $i$ and $j$ are $(0, 0)$, $(1, 2)$ or $(2, 1)$. That gives $3 \times 3 + 2 \times 2 + 2 \times 2 = 17$ cells of colour $0$. Colours $1$ and $2$ have $16$ cells each.

(b) A tile in a row covers $(i, j)$, $(i, j + 1)$ and $(i, j + 2)$. Their sums $i + j$ are three consecutive whole numbers, and three consecutive whole numbers leave the remainders $0$, $1$, $2$ in some order. The same holds for a tile in a column. So the $16$ tiles cover exactly $16$ cells of each colour. Colour $0$ has $17$ cells, so the uncovered cell has colour $0$.

(c) Along a row or a column, $i - j$ also changes by $1$ from one cell to the next, so every tile again covers one cell of each colour. Now colour $0$ means $i$ and $j$ leave the same remainder: $3 \times 3 + 2 \times 2 + 2 \times 2 = 17$ cells, and the other colours have $16$ each. So the uncovered cell has $i - j$ a multiple of $3$ as well as $i + j$. Adding, $2i$ is a multiple of $3$, and since $3$ is prime and does not divide $2$, $i$ is a multiple of $3$. Then $j = (i + j) - i$ is a multiple of $3$ too.

(d) By (c) the uncovered cell has $i$ and $j$ in $\{0, 3, 6\}$, so there are at most $9$ such cells. Each of them works. Corner $(0, 0)$: two tiles cover the rest of row $0$, and the $6 \times 7$ block of rows $1$ to $6$ takes two upright tiles in each column, $14$ tiles. Edge cell $(0, 3)$: two tiles cover cells $(0, 0)$ to $(0, 2)$ and $(0, 4)$ to $(0, 6)$, and rows $1$ to $6$ are covered as before. Centre $(3, 3)$: the rest of the board splits into four $3 \times 4$ rectangles that turn round the centre: rows $0$ to $2$ with columns $0$ to $3$, rows $0$ to $3$ with columns $4$ to $6$, rows $4$ to $6$ with columns $3$ to $6$, and rows $3$ to $6$ with columns $0$ to $2$. Each takes $4$ tiles. Turning or reflecting the board maps $\{0, 3, 6\}$ to itself, so the other corners and edge cells follow. The answer is $9$.

(e) Among $0$ to $9$, four numbers leave remainder $0$ and three leave each of $1$ and $2$. Colour $0$ for $i + j$ has $4 \times 4 + 3 \times 3 + 3 \times 3 = 34$ cells, and the other colours $33$ each. The same count holds for $i - j$. As before, $33$ tiles cover $33$ cells of each colour, so the hole has $i$ and $j$ in $\{0, 3, 6, 9\}$: at most $16$ cells. Each works. Take rows $0$ to $6$ if $i \le 6$ and rows $3$ to $9$ if $i = 9$, and choose seven columns the same way. This $7 \times 7$ block contains the hole at a cell whose coordinates inside the block are in $\{0, 3, 6\}$, so (d) covers the block with $16$ tiles. The rest of the board is a strip three cells wide: a $3 \times 10$ part and a $7 \times 3$ part, covered by $10 + 7 = 17$ tiles laid across the strip. That is $33$ tiles, and the answer is $16$.