So my method for these sorts of things is to start with what I know about the first few moves of the puzzle and what I know about what the last few moves will have to be, and solve from there.

There are two ways to get to 9L: either have 1L of excess (and then all you have to do is add that to the 8L; like I did in my first solution) or have 1L of dearth (and then empty a full 10L until it's full, like I did in the final solution).

Then, starting from the beginning, there are only so many permutations (filling one of the 3 jugs and pouring it into one of the remaining 2 jugs; only 6 ways to start out, really), and just kinda guess which ones seem the most likely to get me to +1 or -1.

So basically, you can 1) fill the 8L first, and your options are to 1a) pour that into the 5L. This gives you 3L remaining, which I noticed very quickly you can get 3L several different ways, so in a puzzle that only has 1 solution, I just assumed it wasn't the best starting place. 1b) You can pour into the 10L, which leaves you 2L of space that you can use to get to 3L (pouring the 5L into it; there's an easier way to get 3L, so I assumed this was worthless) or 6 (pouring the 8L into it again). Hurray, 6L is 1L excess of the 5L bottle, so we have a solution (this was my first solution), but you just can't do it in 8 steps.

2) You can fill the 5L first, and from there you can 2a) fill into the 10L, but that's a worthless move because it just gives you 5L of water (which you can simply get by filling the 5L jug) and 5L of capacity (which you have from the start with the 5L jug). 2b) Fill the 8L, which gives you that number 3 again, which you can get a bunch of different ways. Or so I thought; this is why I had to start from scratch because this is actually a capacity of 3, not an excess of 3, and having a capacity of 3 is unique to this permutation and does in fact get you to 9 quickly.

3) Fill the 10L first, then 3a) fill the 5L jug, which gives you that worthless +5 and -5 situation again (see 2a), or 3b) fill in the 8L and have 2 left over, which if you add to 5, you get 7 which SUCCESS! is one short of 8, so you can solve that way. That's actually a variation on what I did (I just did the 5 -> 8 part first instead of 10 -> 8 part first).

So yeah, I guess just knowing where you're trying to get to and then using a rough process of elimination, and only moving onto the "possible but unlikely" scenarios once all your likely scenarios are exhausted.

Not that anyone's looking for advice on logic puzzles or cares how I do them, but whatever, the post is typed out now, I might as well push enter.