Big Gamma $\Gamma$ meets little gamma $\gamma$
As shown in this answer, $\Gamma'(1)=-\gamma$. Thus, $\Gamma\left(1+\frac1x\right)=1-\frac\gamma{x}+O\left(\frac1{x^2}\right)$ and therefore, $$ x\log\left(\Gamma\left(1+\frac1x\right)\right)=-\gamma+O\left(\frac1x\right) $$ and $$ \lim_{x\to\infty}\Gamma\left(1+\frac1x\right)^{\large x}=e^{-\gamma} $$
The second question is essentially the same as $$ \lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e $$ mentioned in this answer if we set $x=\frac1n$, since $n!=\Gamma(1+n)$.
By Stirling's Approximation, $$ n!\sim\sqrt{2\pi n}\,n^ne^{-n} $$ therefore, $$ \begin{align} \lim_{n\to\infty}\frac1n(n!)^{1/n} &=\lim_{n\to\infty}\frac1n\frac ne\lim_{n\to\infty}\sqrt{2\pi n}^{1/n}\\ &=\frac1e \end{align} $$
First limit: take the logarithm and use that $\gamma=-\psi(1)=-\Gamma'(1)$.
Second limit: take the logarithm and use Stirling's approximation $\ln\Gamma(1+z)=z(\ln z-1)+O(1)$ as $z\rightarrow+\infty$.