Color the grid like a chessboard, with the corners white squares. If you bomb all the black squares, followed by the white squares, followed by the black squares again this guarantees destroying the tank, for a total of 2521 shots. Not sure how to show this is the least.

To show this is best, note that given any pair of adjacent squares $A,B$, your shots must either fire on $B$ then $A$ then $B$ (not necessarily consecutively) or $A$ then $B$ then $A$. This is the only way to account for both possibilities that the tank starts in $A$ and is destroyed on $B$ or starts in $B$ and is destroyed on $A$. In particular, for any pair of adjacent squares, one must be shot twice. The least number of squares you can shoot twice while still hitting every adjacent pair of squares are the black squares in your coloring.

The answer is $\frac{3\cdot 41^2-1}{2}=2521$ Indeed, this follows if we colour the chessboard alternatively (more white squares) and fire at all black squares, then all white ones and then again all black ones. This ensures that we win. Now, otherwise, we can assume that each square has been hit once else there is the possibility of the tank being unharmed at all. Also, each square can't be hit at least $2$ times since that would exceed the obvious bound of $2.41^2$. Thus, there is some square which is hit only once. We mark any such square by an $\times$ and tile the rest of the board with dominoes. In each square we write the number of hits it received. If the sum of numbers in each domino is at least $3$ then we have at least $\frac{3\cdot (41^2-1)}{2}+1$ shots fired. If not, then some domino has sum less or equal to $2$ meaning both of its squares are fired exactly once. Since the pilot doesn't know the position of the tank, we may assume (since the pilot's strategy has to be deterministic) that the tank was placed at the square of the domino which was fired at later and after receiving the first hit, it immediately shifted to the other square of the domino. This proves the result.

The answer is $\frac{3\times 41^2-1}{2}=2521$ Let $k=41^2$. The problem is equivalent to the following: Let $C_1,C_2,...,C_{k}$ be the cells of the $41\times 41$ grid. Suppose the sequence $a_1,a_2,...,a_n$ satisfies the following properties: (i) $1\leq a_i\leq k$ for each $1\leq i\leq n$ (ii) If $C_i$ and $C_j$ are adjacent cells of the grid then both $(i,j)$ and $(j,i)$ are subsequences of $\{a_m\}$ Find the minimum value of $n$. We first provide a construction. We color the grid using chessboard coloring. Suppose all corners are black. Let $C_1,C_2,...,C_{\frac{k-1}{2}}$ are the white cells and $C_{\frac{k+1}{2}},...,C_{k}$ are the black cells. Then since a pair of neighbors must be of different color, the sequence $$1,...,\frac{k-1}{2},\frac{k+1}{2},...,k,1,...,\frac{k-1}{2}$$satisfies the condition, while the length of the sequence is exactly $\frac{3k-1}{2}$ We now prove the bound. From condition (ii) every number in the range $[1,k]$ must appear once in the sequence. We call a number $rare$ if it only appear once. Notice that if there are at least $\frac{k+3}{2}$ rare numbers then by pigeonhole principle there must exist two rare numbers $i,j$ such that $C_i$ and $C_j$ are adjacent. However only one of $(i,j)$ and $(j,i)$ can be a subsequence of the sequence, contradiction. Hence there are at most $\frac{k+1}{2}$ rare numbers. Therefore $$n\geq k+\frac{k+1}{2}=\frac{3k-1}{2}$$

Nice russian combo problem! The answer is $840\cdot3+1=2521$. For a construction, color the grid like a chessboard with corners black. Shoot the $840$ white squares first, then the $841$ black squares, and then the $840$ white squares again, which can be seen to be sufficient. To prove that this is sufficient, partition the square into $840$ $1 \times 2$ rectangles (which can be rotated) and a $1 \times 1$ square. Clearly we must shoot each square at least once, otherwise if the tank lies on some square that we don't shoot, it will remain undestroyed. For a given domino, suppose that we shoot at each of its squares exactly once, and let these squares be $a$ and $b$ in that order. Then if the tank starts at $b$ and moves to $a$ after being hit for the first time, it remains there without getting hit again, which means that $3$ shots are necessary for each domino. This gives the desired bound. $\blacksquare$

Cute russian combo! The answer is $2521$ Do chessboard coloring s.t. there are more white squares, bomb black, then white, and then black. To show that this is optimal, consider two consecutive cells $(A,B)$ , it is easy to see that one must be shot twice, (while the other once). Hence done.

Color the board like a checkerboard with 841 black squares and 840 white squares. Bomb the white squares first, then bomb the black squares, then bomb the white squares. That's 2521 shots. Partition the grid into 840 1 by 2 rectangles of white+black squares and one black square. Every 1 by 2 rectangle must be bombed at least 3 times because otherwise a tank can successfully hide out. The black square left must be bombed at least once because else a tank can just camp there. Thus we need to take at least 2521 shots.