3. 36.

Since a $10 \times 2$ grid has only two cells per row, at most one cell can be marked in each row. Therefore, the nine marked cells must come from nine of the ten rows. Consequently, there must be one row where both cells are unmarked, which we call a "free row."
Consider the following two scenarios:
(1) The first row or the last row is free. Without loss of generality, assume the first row is free. In this case, the second row has two possible ways to be marked. The third row must have one cell marked, and this marked cell must not be in the same column as the marked cell in the second row, so the choice for the third row is determined by the second row. Similarly, the choices for the remaining rows are also determined. Thus, when the first row is free, there are 2 possibilities. Therefore, there are $2 \times 2 = 4$ possibilities in total.
(2) The free row is neither the first nor the last row. In this case, the grid is divided into two parts, and each row in these two parts has exactly one cell marked. Similar to (1), the choices for the cells in the two parts are determined by the first row, giving $2 \times 2 = 4$ possibilities. Therefore, there are $4 \times 8 = 32$ possibilities in total.
In summary, there are $4 + 32 = 36$ possible scenarios.
3. There is a $10 \times 2$ unit square grid. If two cells in the grid share a common edge, they are called "adjacent". Then, there are $\qquad$ ways to mark nine cells such that no two of them are adjacent.