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

$A$ function $f: R \rightarrow R$ is such that $f(1)=2$ and $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. The area (in square units) enclosed by the lines $2|x|+5|y| \leq 4$ expressed in terms of $f(1)$,$f(2)$,and $f(4)$ is

If $f(x + y) = f(x)f(y)$ and $\sum_{x=1}^{\infty} f(x) = 2$,where $x, y \in N$ and $N$ is the set of all natural numbers,then the value of $\frac{f(4)}{f(2)}$ is

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

If $f(x) = \frac{x}{x - 1}$,then $\frac{f(a)}{f(a + 1)} = $

If a function $f$ satisfies $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$ and $f(1) = 1$,then the largest natural number $\lambda$ such that $\sum_{k=1}^{2022} f(\lambda+k) \leq (2022)^2$ is equal to ..........