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 $f(x)$ be defined for all $x > 0$ and be continuous. Let $f(x)$ satisfy $f\left( \frac{x}{y} \right) = f(x) - f(y)$ for all $x, y > 0$ and $f(e) = 1$. Then:

If $f: R \rightarrow R$ is a differentiable function such that $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$ and if $f^{\prime}(4)=24$ and $f^{\prime}(0)=3$,then $f(4)=$

$f(x)$ is a real-valued function such that $2f(x) + 3f(-x) = 15 - 4x$ for all $x \in R$. Then $f(2) =$

Let $f$ be a non-zero real-valued continuous function satisfying $f(x+y) = f(x) \cdot f(y)$ for all $x, y \in R$. If $f(2) = 9$, then $f(6)$ is equal to

If $f(x)$ satisfies the relation $f\left( \frac{5x - 3y}{2} \right) = \frac{5f(x) - 3f(y)}{2}$ for all $x, y \in R$,with $f(0) = 1$ and $f'(0) = 2$,then the period of $\sin(f(x))$ is: