Determine whether each of the functions $2^{n+1}$ and $2^{2n}$ is $O(2^n)$. Determine whether each of the functions $2^{n+1}$ and $2^{2n}$ is $O(2^n)$.
Since $2^n$ < $2^{n+1}$, you can say $2^{n+1}$ is  not $O(2^{n})$
Since $2^n$ is < $2^{2n}$, you can say  $2^{2n}$ is not $O(2^{n})$
Edit: more information:
Since $2^{n+1}$ < $3^{n+1}$ = (n+1) log 3 = O (n log3). 
Since $2^{2n}$ < $3^{2n}$ = O (2n log3) = O (n log3).

You are missing the fact that there must be some constant. The definition is that $f(x)$ is $O(g(x))$ iff there exists some $c$ such that for sufficient large $x$, $|f(x)|\leq c|g(x)|$.
So, a hint for the first one is $2^{n+1}=2*2^n$,
for the second one you might want to check $x^2$ versus $x$. Does $x^2$ belong to $O(x)$?
Hope it helps.
