Some initial deductions:
These are pretty straightforward - first make the X with the 0, then extend dead ends until they have at least 4 cells (or as many as you can make).
In the bottom right,
the 3 joined cells can't use the lower 3, or they trap another cell near the corner. This extends with some "don't repeat the F pentomino" logic to give some more progress:
the 2 clue can't be a dead end of a region. So it must extend up and right, and then it can't be part of the 3 clue. So that places the U in the top right.
And now, there's not much progress that can be made without thinking more globally.
There are 64 cells in the puzzle. There are only 5 tetrominoes available to us, and we need to use enough tetrominoes so that the remainder is a multiple of 5.
The only way to do this is to use exactly one tetromino. This also means we'll need to use all 12 pentominoes.
This lets us resolve the lower right section:
Continuing with this newfound knowledge,
the 3 clue in R5C1 can't go right, because it couldn't make any unused pentominoes. (And it can't be a tetromino in that case, because the bottom left would be an L tetromino.) So that places the L pentomino.
Meanwhile, the I pentomino can only go in the top row.
We now only have the N, Y, W, and T pentominoes left.
Finishing it off:
There's only one place the T pentomino can go: in the upper left corner.
If the 3 connected with the shape below it and made a Y, we wouldn't have anywhere to put the W pentomino.
And finally, there's only one place the Y can go now, and that finishes off the puzzle.
The final answer: