$f(x+y)=f(x)+f(y)+99$, $f(100)=101$, find $f(7)$.

Let $g(x)=f(x)+99$. Then $g(100)=200$ and $$ g(x+y)=f(x+y)+99=f(x)+f(y)+2\cdot 99 = g(x)+g(y).$$ With $x=y=0$, find $g(0)=0$. Then note that $g(n+1)=g(n)+g(1)$ and conclude $g(n)=ng(1)$ for $n\in\Bbb N$.

Suppose $f(1)=A$.

Then: $$f(2)=2f(1)+99=2A+99$$ $$f(3)=f(2)+f(1)=3A+198$$ $$f(4)=f(1)+f(3)+99=4A+3(99)$$

We are given the general rule that for all $n\in\mathbb{Z}$, $f(n)=nA+99(n-1)$. We can plug in $n=100$ and get that $101=100A+99^2$, which gives that $A=-97$. Then, substituting $n=7$, we have that $f(7)=7(-97)+6\cdot99=-85$