For $E$ over $F$ a Galois extension with $G=\mathrm{Gal}(E/F)$. Prove that for certain $x∈E, x∉F$, the sum of $\phi(x)$ among all $\phi∈G$ is in F, and the product of $\phi(x)$ among all $\phi∈G$ is in F as well. And also prove that any $y∈F$ can be expressed as the sum of $\phi(x)$ for some $x$.
I can understand this by some examples, like $\mathrm{Gal}(\mathbb Q(\sqrt2)/\mathbb Q)$, then the sum of $\phi(a+b\sqrt2)$ is just $(a+b\sqrt2)+(a-b\sqrt2)=2a$, so the conjugate parts always get canceled and the result lies in $\mathbb Q$. And for a given $y$ we can make $a=\frac y2$ and get what we need. But I can't extend these results to any Galois extension $E$ over $F$.
This is an example where it is (perhaps) easier to see what is going on abstractly instead of working with particular examples. Let
$$\mathrm{Tr}(x) = \sum_{\sigma \in G} \sigma(x), N(x) = \prod_{\sigma \in G} \sigma(x)$$
To show that $\mathrm{Tr}(x) \in F$ for all $x \in E$, it is sufficient to show that every element of $G$ fixes $\mathrm{Tr}(x)$. To that end, note that for any $\tau \in G$, we have
$$\tau(\mathrm{Tr}(x)) = \tau\left(\sum_{\sigma \in G} \sigma(x)\right) = \sum_{\sigma \in G} \tau\sigma(x) = \sum_{\gamma \in \tau G} \gamma(x) = \sum_{\gamma \in G} \gamma(x) = \mathrm{Tr}(x)$$ where the second to last equality holds because $\tau G = G$; that is, left-multiplication by $\tau$ induces a bijection $G \to G$ (whose inverse is left-multiplication by $\tau^{-1}$).
I encourage you to try to write a similar argument for $N(x)$.