Answer
$f(2) =6$
$f(3)= 9$
$f(4) =12.$
No.
Work Step by Step
$f(2)=f(1+1)=$ by definition of f$ =f(1)+f(1)=3+3=6$
$f(3)=f(1+2)=$ by definition of f$ =f(1)+f(2)=3+6=9$
$f(4)=f(1+3)=$ by definition of f$ =f(1)+f(3)=3+9=12$
For the second question, take $f(x)=x^{2}$
$f(2)=4,$
$f(3)=9$
$f(2)+f(3)=4+9=13$
$f(2+3)=f(5)=25\neq 13$
So, no.
