The completed grid:

Step-by-step solution:
As a first step, note that for all the columns and rows with a gap of size at least 6, we know that some of the middle cells need to be empty:

This yields a number of other cells which also need to be empty, since they cannot form a gap of the required size:

Now comes the first deduction which is a bit trickier:
Consider the two rows with the 2 and the 5. In the case that the 5-row does not have a star in the first or last column of the grid, then it will block off cells in the 2-row above it in a specific pattern: the first few (0, 1, 2 or 3) cells are not blocked off, then 3 cells are blocked off, then 3 cells are not blocked off again, and finally a few (0, 1, 2 or 3) cells are not blocked off again. For example, if the 5-row has stars at positions 3 and 9, then the pattern of blocked off cells looks as follows (focus on the row with the 2):

Note that in such a situation, it is impossible to complete the 2 row in a valid manner. The only way of avoiding a situation like this, is by filling the 5 row as follows:

This leaves only one way of completing the 2 row. By furthermore marking some cells which cannot contain stars anymore, we get the following situation:

Next, we look at
the consecutive columns with the 1 and 6. These two columns can only be filled in a single manner: The 6-colums will have a star somewhere in the top four cells, which makes it impossible for the stars of the 1-column to go in the top four cells. This 1-column should therefore have a star in row 10, which in turn only leaves one possibility for the 6-column:

Now the 9-row can be completed in only one way, which in turn forces the position of one of the stars in the 4-column:

Now we are almost done!
The remaining 6-column (column 7) may not block off both possibilities for completing the 3-column next to it. This eliminates one of the possibilities of this column:

This leaves only one way of filling the 1-row, which leaves one possibility for completing the 4-column:

Now row 2 and column 11 can be completed, which quickly resolves the rest of the puzzle:
