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 $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. Suppose that $f(3)=3$ and $f^{\prime}(0)=11$,then $f^{\prime}(3)$ is given by

If $f(x) + 2f(1/x) = 3x$ for $x \neq 0$ and $S = \{x \in R : f(x) = f(-x)\}$,then $S$:

If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$

Let $f$ be a function defined on the set of all positive integers such that $f(xy) = f(x) + f(y)$ for all positive integers $x, y$. If $f(12) = 24$ and $f(8) = 15$,then the value of $f(48)$ is:

Find $\sum_{t=1}^{39} f(t)$ if $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$.