How to solve the recurrence relation $f(1,n)=n+2$? Given the following information:
$f(0,n)= n+1$ $ \ \ $ $\forall n$
$f(m,0)=f(m-1,1)$ when $m>0$
$f(m,n)=f(m-1, f(m,n-1))$ when $m>0$ and $n>0$ 
I have worked out that $f(1,n)=f(0,f(1,n-1))= f(1,n-1) + 1$
But I am unsure of how to get $f(1,n) =n+2$ from this stage ? 

You already have:
$$f(1,n)=f(1,n-1) + 1$$
Notice $f(0,1)=f(1,0)=2$, and this is because
$f(0,n)= n+1$, and you let $n=1$, you get $f(0,1)= 2$
$f(m,0)=f(m-1,1)$, and you let $m=1$, you get $f(1,0)=f(0,1)$
Thus $f(1,0)=f(0,1)=2$
Let $a_n=f(1,n)$
So you have $a_n=a_{n-1}+1$, and $a_0=2$
Then
$$a_n=(a_n-a_{n-1}) + (a_{n-1} - a_{n-2}) + \ldots +(a_1-a_0)+a_0=n+ 2$$
