I think you forgot this line: Determine f(1991) and the most general form of g.

For the RHS of the equation, it is $ -f(f(m+1))-n$ or $ -f(f(m+1)-n)$.

the former

As we can choose $ n$ freely, so the function is surjective. So there exists a constant integer $ k$ such that $ f(f(k))=1$, and we have $ f(m+1)=-f(f(m+1))-k$. Again, by surjectivity, for all integer $ n$, we have $ n=-f(n)-k$, i.e. $ f(n)=-n-k$. Substitute this into (a), $ -m-n-k=-m-1-n$, and $ k=1$. So $ f(1991)=-1991-1=-1992$.