How many functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ are there such that $f(x+y)=f(x)+f(y)$ for all $x, y \in \mathbb{Z}$?

Let $f$ be a function such that $f(x + y) = f(x) + f(y)$ for all $x$ and $y$,and $f(x) = (2x^2 + 3x)g(x)$ for all $x$,where $g(x)$ is continuous and $g(0) = 3$. Then $f'(x)$ is equal to:

Let $f: N \rightarrow N$ be a function such that $f(m+n)=f(m)+f(n)$ for every $m, n \in N$. If $f(6)=18$ then $f(2) \cdot f(3)$ is equal to :

$f: R \rightarrow R$ is a function such that $f(0)=1$ and for all $x, y \in R$,$f(xy+1)=f(x)f(y)-f(y)-x+2$. Then $\frac{df}{dx}$ at $x=e$ is:

Let $f$ and $g$ be functions satisfying $f(x+y)=f(x)f(y)$,$f(1)=7$ and $g(x+y)=g(xy)$,$g(1)=1$ for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$,then $n$ is equal to:

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y.$ If $f(30) = 20,$ then the value of $f(40)$ is-