If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x = 1}^n {f\left( x \right) = 120,} $ find the value of $n$

Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3 .$ If $\sum \limits_{i=1}^{n} f(i)=363,$ then $n$ is equal to

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

Greatest value of the function, $f(x) =  – 1 + \frac{2}{{{2^x}^2 + 1}}$ is