For most positions of the holes there seems to be room for 10 rooks.
If the holes are in the same row (not on the edge, not touching each other, and with at least two rows above and below it), then place 7 rooks as
R
R
R o R o R
R
R
There are then 3 rows and 3 columns left without anything in them yet; put the last 3 rooks there.
If the holes are in different rows and columns (but not on the edge of the board), place 6 rooks as
R
R o R
R o R
R
There will be 4 rows and 4 columns yet unused; place the 4 last rooks there.
These solutions are clearly optimal for their hole placements, because each maximal horizontal or vertical run of available cells on the board contains a rook.
If the holes are too close to the edges (or to each other) for this to work, there's room for fewer rooks. Writing down exact rules for those cases seems to be tedious but doable.
If you just want an algorithm to find the maximum number of rooks rather than actually place them, one way that seems to work is to count the number of different horizontal hole-less runs of cells on the board, and the number of different vertical hole-less runs of cells on the board. The minimum of these two numbers is obviuously an upper bound on the rooks, and it can always be achieved (this I can only show with a tedious case analysis).
So for the first question you need to place a rook in each column. The rook in the first column can go in $8$ places, the rook in the second column then has $7$ possible rows etc.
In the second part, you need first to choose the $8$ columns out of the $10$ available, and then place a rook in each of the columns. If once again you move from left to right, counting the number of possibilities, you should be fine.
Rather than talking about "any particular" column or row, you should make a habit of defining the scheme more closely (the first, second rather than "any" etc). This will help to avoid a great deal of confusion and can also highlight any double-counting.
Best Answer
For bishops choose a colour - in one direction you will find seven diagonals. So at most one on each diagonal, two colours, fourteen bishops - realised by eight on the first rank and six on the eighth.
For rooks, one on each rank - eight along the diagonal will do.
For knights - thirty two all on the same colour.
Not sure what you mean about pawns.
This on a standard chessboard, but generalisable. If the side of the chessboard is odd, put the knights on the colour with most squares.