I am confident that both functions f and g have inverse functions, because they are one-to-one and onto. But I am having trouble proving it.
When I draw the image for f((3,4)), on the left hand side (3,4) points to (4,3) (on RHS) and on the left hand side 4 points to 3 (on RHS). When I draw the inverse function, (4,3) points to (3,4). Each point on the image on the right hand side only has one arrow pointing into it from the left hand side, which is why I believe that it is one-to-one.
The image of f is equal to its domain (3,4) = (3,4) hence I believe that it is onto.
Does my logic and proof sound correct? I have the feeling that it could be explained in a more concrete manner.
Your idea on the first one is correct, for the second one it's not so easy. Note that if $g(x,y) = (3,4)$ we clearly have $y=4$ but $x$ could be both $6$ and $-6$. If the domain of $g$ was $\mathbb{Z}^+\times \mathbb{Z}$ it would be a different story, but as stated, is $g^{-1}$ a function?