Given
\begin{aligned} f(n)&=\frac{n^2+2}{\; 1+2^{-n}}\\[1em] g(n)&=\ \ n+3 \end{aligned} I have to calculate the constant $c$ and the $n_0$ of $g(n)=O(f(n))$.
$$n+3≤n^2+3≤3n^2+3≤3n^2+6≤3(n^2+2)$$
At this point I don't know how to go on because if I divide $3(n^2+2)$ by something I get something less than $3(n^2+2)$.
HINT of a possible method: Note that $$\frac{1}{1+2^{-n}} \ge \frac{1}{1+ 2^{-1}} = \frac{2}{3}$$ for all $n \ge 1$. So $f(n) \ge \frac{2}{3}\left(n^2+2\right)$ for all $n\ge 1$. Therefore, it is sufficient for you to find $c$ and $n_0$ such that $$g(n) \le c \times \frac{2}{3}\left(n^2+2\right)$$ for all positive integers $n\ge n_0$. Are you able to do this (note: you have actually already done most of the work for it!)?