Failure of Stone-Weierstrass theorem on infinite domain

Basically by definition, the sup norm is continuous with respect to the topology of uniform convergence. Since $\|\sin x\|_{\infty}=1$ and every non-constant polynomial $p(x)$ satisfies $\|p(x)\|_{\infty}=\infty$, it follows immediately that $\sin x$ admits no uniform approximation by polynomials on $\mathbb R$.

For $e^x$ the same argument applies after restricting the domain to $(-\infty,0]$, since any uniform approximation on $\mathbb R$ will also restrict to a uniform approximation on $(-\infty,0]$.

For $e^{x}$ simply use the fact that it remains bounded as $x \to -\infty$ whereas the absolute value of a polynomial tends to $\infty$.

For the first function your answer is correct.