Number of permutations of BANANA if (1) the two Ns are separated and (2) the three As are all separated One way to do (1) is subtract all permutations of BAXAA (where X = NN) from all the permutations of BANANA: $\frac{6!}{2!3!} - \frac{5!}{3!}.$
I was wondering if it is possible to solve (1) by letting X = NBN, Y = NAN, counting the number of permutations of BAYA and XAAA and taking their sum to get $\frac{4!}{3!} + 3(\frac{4!}{2!})$ where the coefficient $3$ is to account for three different A's. 
The reason I am unsure about the second method for (1) is that I can't seem to get (2) right. Here I let X = AA, Y = AAA, count the number of permutations of BXNNA and BYNN and subtract their sum from the total number of permutations of BANANA which is $\frac{6!}{2!3!} - (\frac{4!}{2!} + \frac{5!}{2!}) = -12.$
My questions.
Is the second method for (1) correct? Why do I keep getting $-12$ in the solution of (2)? How do I fix it? 
 A: 
Is the second method for (1) Correct?

For (1), your first method is correct, your second is correct for the wrong reasons.  Notice that you can have the $N$'s be separated by more than one letter.  You do not count $NBAAAN$ for example in your count.  On the other hand you inflated the number by multiplying by three when you shouldn't have... the $A$'s are all identical, so we don't care which of the $A$'s was sandwiched between the $N$'s since they result in the same outcomes.
If you were to approach directly, you'd have to take into account not only the case with $NAN$ and $NBN$, but also $NAAN, NABN, NBAN, NAAAN, NAABN,\dots$ and is entirely too much work.

Why do I keep getting $-12$ in the solution of (2)?

Some of the arrangements of $BXNNA$ and $BYNN$ are the same, for example $BXANN$, $BAXNN$ and $BYNN$ are all referring to $BAAANN$ but you subtracted it from the count three times when you should have only subtracted it once.

How do I fix it?

For a correct solution to part (2), first pick an arrangement of $XXXAAA$ such that no $A$'s are next to one another and then afterwards replace the $X$'s by an arrangement of $BNN$


*

*Arranging $XXXAAA$ such that no two $A$'s are adjacent:


You may do this more formally using stars and bars (which you should do for more complicated problems) or you may simply brute force count the number of valid arrangements by hand.  They are:
$AXAXAX, AXAXXA, AXXAXA, XAXAXA$


*

*Replacing the $X$'s in whichever of the selected arrangements from the previous step by an arrangement of $BB$


Pick which of the three locations currently occupied by an $X$ is to be taken by the $B$.  The remaining locations are then used by the $N$'s.
Multiplying, there are then $4\cdot 3=12$ valid arrangements.
