Player Zero has the following strategy: do not place a 0 in the first two columns unless you have no other choice. In that case, place a 0 in a row where the other cell of the first two cells already contains a 1, if possible. If such a cell does not exist, place it in a row that does not yet have a 0 in the first two cells. We will show that Zero can continue to do this all the time, so that Zero wins because there will never be a row that starts with two 0s.
We claim that with this strategy, at any moment, there is at most one row whose first two cells are $(0, \emptyset)$. Suppose that Zero places a zero next to an empty cell. Then the right area is full due to the strategy. When One makes a move after this, it replaces the $(0, \emptyset)$-row with a $(0,1)$-row, or a $(1, \emptyset)$-row is created where Zero can/must play. Zero is thus never forced to create a second $(0, \emptyset)$-row, or a $(0,0)$-row.
b) Solution. Zero can only ensure that no row is divisible by 3 if $n \equiv 2 \bmod 4$.

First note that modulo 3 the columns are numbered $1,2,1,2, \ldots$. If $n$ is even, there are an even number of cells and One has the last turn; if $n$ is odd, there are an odd number of cells and Zero has the last turn.

If $n=4k$, both 1 and 2 appear an even number of times per row. Now One has the following strategy: mirror player Zero. If Zero places a zero in a cell with value 1, then One places a 1 in a cell in the same row that also has value 1. If Zero places a zero in a cell with value 2, then One places a 1 in a cell with value 2 in the same row. Since there is an even number of 1s and 2s in each row, One can always continue to do this. Now it holds that at the end $k$ cells with value 1 have the number 1 and $k$ cells with value 2 have the number 1 from One. So the sum of each row is $3k \equiv 0 \bmod 3$.

If $n=4k+1$, there are $2k+1$ cells with value 1 and $2k$ cells with value 2. One wants to create a row such that exactly $k$ cells with value 1 have a 1 and exactly $k$ cells with value 2. One now uses the mirror strategy again. However, if One cannot mirror, then One places a 1 in any cell with value 1. Only if this is also no longer possible does he place a 1 in a cell with value 2. Since there are exactly $2k$ cells per row with value 2, One can ensure with this strategy that in each row exactly $k$ cells with value 2 get the number 1. In addition, this strategy ensures that there is at least one row with $k+1$ zeros in cells with value 1 (and thus $k$ with the number 1). For this row, the sum is $k+2k=3k \equiv 0 \bmod 3$.

If $n=4k+3$, there are $2k+2$ cells with value 1 and $2k+1$ cells with value 2. One starts by placing a 1 in any cell with value 2 in one of the rows where Zero has not placed his first cell. One then applies the copy technique to this specific row. So if Zero places it outside, One does the same and if Zero places it inside, One also does this and places his 1 on a cell with the same value. Since there is now an odd number of empty cells outside this row, it can happen that One has to place a cell in the special row while Zero did not do so on the previous turn. There are still $2k$ cells with value 2 and $2k+2$ with value 1 for this move. By applying the copy technique again after this move (and the fact that Zero has the last move), One can ensure that both One and Zero choose $k$ cells with value 2 and $k+1$ cells with value 1. Now the special row has exactly $k+1$ cells with value 1 and $k+1$ with value 2, so the sum is $3(k+1) \equiv 0 \bmod 3$ and One wins.

If $n=4k+2$, there are $2k+1$ cells with value 1 and $2k+1$ cells with value 2. Then Zero can prevent One from winning. This time it is Zero who applies the copy technique from his second turn. However, if there are no more cells with the same value in the same row, then Zero places the cell in the same row with the other value. Since in this case One has the last turn, Zero can ensure with this strategy that both exactly $2k+1$ cells in each row are occupied and that Zero occupies at least $k$ cells with value 1 and $k$ cells with value 2 in the row. This means that the sum in each row is either $k+2(k+1) \equiv 2 \bmod 3$ or $k+1+2k \equiv 1 \bmod 3$. In this case, Zero can always prevent One from winning.