(1) I have to prove the following: $\forall p \in \mathbb{P} \setminus \left\{ {2}\right\}, \sum_{k=1}^{p-1}(\zeta_{p}^{k})=-1$ where $\mathbb{P}=\left\{ {p\in \mathbb{Z}:p>0, prime}\right\}$.
Intuitively, it makes sense. (I even understand the exception for $p=2$). Basically, $\zeta_{p}^{p} = 1$ is the only term missing ($mod{p}$ in the power) and the imaginary parts and real parts of the rest each cancel to $0$ except for $-1$ (because of the missing term). I can see it pictorially. But showing it algebraically is a different story.
(2) I also have to find: $\prod_{k=1}^{p-1}(1-\zeta_{p}^{k})$ and $\prod_{k\in\mathbb{Z}^{+}\cap[1,p-2]:k\equiv1 (mod{2})}((\zeta_{p}^{k}-\zeta_{p}^{-k})^{2})$.
I am not sure how to work with these. Can you provide an answer with justification to each?
I see that the second one looks like $sin$ but I am not sure how helpful that is.
(3) How do I show that $\exists$ field $\mathbb{F}: \mathbb{Q} \subseteq \mathbb{F} \subseteq \mathbb{Q}(\zeta_{p}), [\mathbb{F}:\mathbb{Q}]=2$? What is the square-free $\alpha: \mathbb{F}=\mathbb{Q}(\sqrt{\alpha}), (\nexists q \in \mathbb{Q} \setminus \{1\}: q^2=\alpha)$?
I can give you a couple of hints.
For $(1)$, notice that your sum of roots of unity is a geometric sum of the form $$\sum_{k=1}^{p-1} z^k$$
For $(2)$, since all the $p$-th roots of unity have the form $\zeta_p^k$ for $0 \leq k \leq p-1$, then the polynomial $z^{p} - 1$ factors as
$$ z^{p} - 1 = \prod_{k=0}^{p-1} (z - \zeta_p^k) $$ but also $z^p -1 = (z -1)(z^{p-1} + z^{p-2} + \cdots + z + 1)$. From this you should be able to evaluate the first product in $(2)$.
For $(3)$, one way of producing explicitly the "square-free $\alpha$" is to evaluate the quadratic Gauss sum
$$ \sum_{k=0}^{p-1} \zeta_p^{k^2} = \begin{cases} \sqrt{p} \quad \text{if $p \equiv 1 \pmod{4}$}\\ \sqrt{-p} \quad \text{if $p \equiv 3 \pmod{4}$} \end{cases} $$ You can easily find how to evaluate this sum online.
Added in response to comments
In $(1)$, what you have is a geometric sum. The following formula for such a sum is (or should be) well known from elementary calculus:
$$ 1 + z + z^2 + \cdots + z^{n} = \sum_{k=0}^n z^k = \frac{z^{n+1}-1}{z-1} $$ Using this, you can derive the result that you have to prove in part $(1)$.