2. Not sure if there is a formal proof of this, but one can see that as $x \to \infty$ then $p(x) \to \infty$ and as $x \to -\infty$ then $p(x) \to -\infty$. (assuming leading coefficient is positive, if negative its just the other way around)
By definition any polynomial is continuous on $\mathbb{R}$ and thus by the IVT we can say there exists at least one $x \in \mathbb{R}$ such that $p(x)=0$