We can write ,
$f(n)=f(n-1)+g(n-1) $ , ………..(1)
$f(0)=1$
$g(0)=1$
$f(1)=f(0)+g(0)=1+1=2$
$g(n)=g(n-1)-f(n)$ , $g(0)=1$
$g(n)=g(n-1)-f(n)$
$g(n)=f(n)-f(n-1)-f(n)$ [as ,from (1) we get $g(n-1)=f(n)-f(n-1)$]
$g(n)=-f(n-1)$ ; $n\geq 1$
$g(n-1)=-f(n-2)$ ; ………….(2)
Putting this in (1) we get ,
$f(n)=f(n-1)-f(n-2)$ $n\geq 2$
$f(0)=1$, $f(1)=2$
$f(2)=f(1)-f(0)=2-1=1$
$f(3)=f(2)-f(1)=1-2=-1$
$g(1)=-f(0)=-1$
$g(2)=-f(1)=-2$
$g(3)=-f(2)=-1$
$g(4)=-f(3)=1$
Now if you checked all the option are correct.