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

If $f:R \to R$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 7$,then $\sum_{r = 1}^n f(r)$ is

Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

Let $f : R \rightarrow R$ be a continuous function such that $f(3x) - f(x) = x$. If $f(8) = 7$,then $f(14)$ is equal to.

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 $f : R - \{0, 1\} \rightarrow R$ be a function such that $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. Then $f(2)$ is equal to: