From options, it is clear that, we need recurrence for $f$ in $g,$ $g$ in $f,$ $f$ in $g$ and $f$ in $f.$

As you have represented $g$ in $f$ in equation $(2)$ we can do it for all above as:

$f(n) = f(n-1) + g(n-1), \ n\geq 1$ $\ \ (1)$

$g(n) = g(n-1) - f(n), \ n\geq 1 \implies g(n-1) = g(n-2) - f(n-1)$

Putting this $g(n-1) = g(n-2) - f(n-1)$ in equation $(1),$ we get,

$f(n) = f(n-1) + g(n-2) - f(n-1)$

So, $f(n) = g(n-2); \ \ n \geq 2$ $\ \ (2)$

Since, you have already proved that $g(n) = -f(n-1) \implies g(n-2) = -f(n-3)$

Now, putting $g(n-2) = -f(n-3)$ in equation $(2),$ we get,

$f(n)= -f(n-3)$ $\ \ (3)$

Since, you have already proved that $g(n) = -f(n-1)$ and since, from equation $(2), f(n-1) = g(n-3)$

So, $g(n) =\ – g(n-3)$

So, our recurrences are:

- $f(n) = g(n-2);\ n\geq 2$
- $g(n) = -f(n-1) ; n \geq 1$
- $f(n) = -f(n-3); \ n\geq 3 $
- $g(n) = -g(n-3); \ n \geq 3$