For (2), suppose first that $k=2m$ is even. Then $\frac{k!}{2^m}=\frac{\big(1\cdot3\cdot5\cdot\ldots\cdot(2m-1)\big)(2\cdot4\cdot6\cdot\ldots\cdot 2m)}{2^m}=\big(1\cdot3\cdot5\cdot\ldots\cdot(2m-1)\big)m!\;,$ so $\frac{2^{1+k/2}}{k!}=\frac2{\big(1\cdot3\cdot5\cdot\ldots\cdot(2m-1)\big)m!}<\frac2{(k/2)!^2}\;,$ which is indeed much smaller than $1$ when $k$ is large. Only minor adjustments are required when $k$ is odd; I’ll leave them to you.
(1) is trivial as written: $0\le\lfloor 2^{k/2}\rfloor\le 2^{k/2}$ by the definition of floor and the fact that $2^{k/2}>0$, so $0\le\frac{\lfloor 2^{k/2}\rfloor}{2^{k/2}}\le 1\;,$ and therefore $0\le\left(\frac{\lfloor 2^{k/2}\rfloor}{2^{k/2}}\right)^k\le 1\;.$
If instead $n=2^{\lfloor k/2\rfloor}$, we have $0<2^{\lfloor k/2\rfloor}\le2^{k/2}$, and the same reasoning applies.