Prove $1 + x + x^2 + x^3 + x^4 > 0$

Solving all parts at once:

As has already been mentioned, $1+x+\dots+x^n=\frac{1-x^{n+1}}{1-x}$ comes in handy here.

Let $f_n(x)=1+x+\dots+x^n$. Then if $x \neq 1$, $f_n(x)=\frac{1-x^{n+1}}{1-x}$

$f_n(x)=0\implies$Clearly $f_n(1)\neq0$, so $1-x^{n+1}=0$

$$\implies x^{n+1}=1$$

$n$ even $\implies x=1$. But $f_n(1)\neq0$, so therefore there aren't any zeroes. Since $f_n(1)>0$, $f_n(x)$ is positive for all $x$.

$n$ odd $\implies x=\pm 1 \implies x=-1$. Verifying: $f_n(-1)=0$, and therefore it is only zero.

What about $x^4+x^3+x^2+x+1 = x^2\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(x+\dfrac{2}{3}\right)^2+\dfrac{2}{3}\geq\dfrac{2}{3}>0.$

Update: Indeed, one can use the similar idea to bound $$f_n(x) = x^{2n}+x^{2n-1}+...+x^2+x+1 =x^{2n-2}\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}x^{2n-4}\left(x+\dfrac{2}{3}\right)^2+$$$$+\dfrac{4}{6}x^{2n-6}\left(x+\dfrac{3}{4}\right)^2+\dots+ \dfrac{n+1}{2n}\left(x+\dfrac{n}{n+1}\right)^2+\dfrac{n+2}{2n+2}> 0.$$ Proving this is a straightforward induction on the sequence $$a_n = \dfrac{n+1}{2n},$$ which happens to satisfy the recurrence relation: $$a_{n+1} = 1-\dfrac{1}{4a_n}.$$ Therefore, we get the lower bound $f_n(x)>\dfrac{n+2}{2n+2},$ which I think is still not that close to the actual minimum value.

Use the geometric series formula: $$ 1+x+\dots+x^n=\frac{1-x^{n+1}}{1-x} $$

If $n$ is even, show that $1-x^{n+1}$ has the same sign as $1-x$ for all nonzero $x$; since $1+x+\dots+x^n$ is clearly positive at $0$, this is enough to prove c) and therefore b).

If $n$ is odd, note that $1-x^{n+1}$ has only two real zeroes, namely $\pm 1$. Since $1+x+\dots+x^n$ is nonzero for $x=1$, this is enough to prove d).