1. What is the minimum number of cells that need to be marked in a $9 \times 9$ table so that each vertical or horizontal strip $1 \times 5$ contains at least one marked cell?
Answer: 16.

Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row and $m$ axial cells in the central column. In rows and columns that do not contain axial cells, there should be at least two marked cells. In total, in such rows, there will be at least $2 \cdot(2 n-2-m)=4(n-1)-2 m$ marked cells, and in columns - at least $4(n-1)-2 k$ cells. At the same time, each cell will be counted no more than twice. Taking into account the axial cells, we get that there are at least

$$
\frac{1}{2}(4(n-1)-2 m+4(n-1)-2 k)+k+m=4(n-1) \text { cells. }
$$

The value $4(n-1)$ is achieved if all cells of the central row and column, except for the center of the board, are marked. Assuming $n=5$, we get the answer.