Quick notes on notation
This is a blank square. We have not deduced anything yet.
This is an unshaded square.
This is a shaded square.
I make use of both an absolute coordinate system... RxCyLz, standing for Row x, Column y, Layer z.
To refer to multiple cells (usually to mark some 2x2's) I parenthesize squares to group them together. So R(2-3)C3L(3-4) refers to the four squares R2C3L3, R2C3L4, R3C3L3, R3C3L4.
I also make use of a relative coordinate system. North, south, east and west always refer to staying in the same layer. Going up is going towards layer 1; going down is going towards layer 4.
Puzzle 1 full solution
Puzzle start. First, the standard Nurikabe deduction of having two diagonally adjacent clues still applies. But in the 3D version, some of these are harder to spot!
The 9 must go north, and then up.
Then north again. The bottom layer 4-clue must go north.
The shaded cells in the top right needs to escape north, and then up.
R4C4L3 must escape to R4C4L2. R3C4L2 must also be shaded to separate R3C4L1 and R3C4L3.
Top layer 4-clue must resolve as follows. It can't go down because it's blocked by shaded squares, and at R4C3L1, going down would result in touching the 3rd layer 4-clue.
Now we surround the top 4-clue.
The 3rd layer 4-clue must continue west to R4C2L3, and the 9-clue must continue up to R2C4L2.
Also, R3C3L2 must be unshaded to avoid creating a 2x2x1.
However, R2C3L2 must be shaded, because otherwise it would make the 2-clue be at least size 3.
We get two more extensions in layer 2 as a result.
R3C2L4 is unshaded or else we make a 2x2x1 from R3C(2-3)L(3-4). Tricky to see, eh?
Extend the isolated shaded group in R1C4L3 to R1C3L3.
Now we can complete the 4th layer 4.
And fill in adjacent squares.
Now, the 9-clue is now 6 long. Its only squares of freedom starts in R1C4L2, where it can go either up or west. Up only has room for 1 square, so it must go west at least twice.
This makes R2C4L1 shaded pretty isolated and it must escape that way.
This finishes up the 2-clue. Also the R3C2L1 must escape west.
Finish the 9 clue...
At this point the two unshaded squares in R3C2-3L2 must go to the 6 clue, in only one way.
And the 4 connects to R3C2L4 like that.
Puzzle 2 full solution
1 clues are great and give us lots of information.
Also some squares clues are separated apart by a block.
Already, layer 3 needs R2C2 to be unshaded to avoid the 2x2.
That completes the 2-clue.
4-clue only has 1 room for expansion going south; so it must go east and down.
R2C4L1 gets shaded to separate. Also, R2C1L4's 3 clue must go up as its only expansion.
Now, to employ some tricky logic. One of R3C1-2L1 must be unshaded, or else we make a 2x2 (R3C(1-2)L(1-2)). Either the 3 extends south, or the 6 gets there, but the 6 would exhaust all of its length doing so. (west, west, up, up, north, say.) That would leave R(2-3)C4L3 unattended (The 4-clue requires 5 to get there; other clues are more hopeless). So the 6 must care for that (we can't immediately mark R3C4L3 as unshaded, because we could conceivably satisfy it by first going down, north, north, up). But we can declare that 3 must go south, south.
Fill adjacents, then note that R2C3L1 must be unshaded to avoid a 2x2. That completes the 4-clue...
Fill those adjacents, note that R1C1L3 must also be unshaded to avoid 2x2's, which completes the R2C1L4 3 clue...
And fill yet more adjacents. Whew.
Now we should look at layer 2's R4C(1-2). One of these must be unshaded, and it must be satisfied by the 6-clue. It must hit R4C2L2 first, so let's path that.
Actually the 3-clue in R3C4L1 is pretty trapped and has only one way to resolve so let's do that.
Fill adjacents. R2C3L4 is too far from the 6-clue now, as it's already size 4. So it completes the size 3 region as follows:
Nothing can reach R1C4L4. Let's shade it.
That means R2C4L3 must be unshaded to avoid another 2x2 (R(1-2)C4L(3-4)) and that completes the 6-clue. We're done!
Puzzle 3 full solution
Let's do the spacing deductions again, and fill out all of the 1's.
We use another 2D nurikabe deduction that often appears with 2's - if a 2-clue only has two places to expand, then the diagonally adjacent square that is adjacent to both of these places is shaded. Otherwise the 2-clue is trapped to either being a 1 or 3+ clue. This shades R2C3L4, since the 2-clue only has 2 places to go now.
R1C1L1's 5 cannot go down at all! This finishes the 5-clue.
And fill adjacents.
Both the 9 clue and above it R1C4L1 must go south to escape. R4C1L1 must go east to escape. This also shades R2C3L2.
Now R3C2L2 must be unshaded to avoid the 2x2 in R(3-4)C2L(1-2). That completes that 2-clue, and fill adjacents.
The 5-clue in R3C3L1 must go down at some point. The only accessible place is R4C4L2.
After shading in R4C3L2, R4C4L3 (both blocked by the 2-clue in R4C3L3), and R3L4C2 (blocked by the 9-clue), this completes the 5-clue.
9-clue must go down. This allows us to fill R2C4L4 (blocked by the 2-clue in R1C4L4)...
...which resolves the 2-clue that way.
The 9-clue must keep going down, however. Only place to do that is via R3C4L4.
Thus R4C4L3 must escape down, west.
9-clue continues westward.
Shade R3C3L3; that resolves the last 2-clue.
R4C3L4 must keep escaping west, west.
R2C3L3 belongs to the 9-clue or else R(2-3)C(2-3)L3 is a 2x2 shaded
R2C3L4 escapes west...
Then the 9-clue escapes west, completing it. Fill adjacents...
Everything else goes to the 4-clue. We're done!