Let $L/K$ be a finite extension and let $Tr_{L/K}$ $N_{L/K}$ be its trace and norm of the extension $L/K$ Let $\pi_{\alpha ,K}(X) $ be the minimal polynomial of $\alpha$ over $K$ which is split completely as $(X-\alpha_{1})...(X-\alpha_{d})$ over a large enough field extension .then
1)$Tr_{L/K}(\alpha)=[L:K(\alpha)](\alpha_{1}+...+\alpha_{d})$
2)$N_{L/K}(\alpha)=(\alpha_{1}....\alpha_{d})^{[L:K(\alpha)]}$
If we write the minimal polynomial as $\pi _{\alpha}(X)=x^{d}+c_{d-1}X^{d-1}+...+c_{0}.$ then
3)$Tr_{L/K}(\alpha)=-[L:K(\alpha)]c_{d-1}$
4)$N_{L/K}(\alpha)=(-1)^{n}c_{0}^{[L:K(\alpha)]}$
Now let assume that the extension $L/K$ is Galois with a Galois group
$G=Gal(L/K)$ here we define the trace and the norm as
$Tr_{L/K}(\alpha)=\displaystyle \sum_{\sigma \in G } \sigma(\alpha) $, $ N_{L/K}(\alpha)=\displaystyle\prod_{\sigma \in G }\sigma(\alpha)$
I ask whether the property above of a trace and norm are still correct in the case of Galois extension?
The extension $L|K$ is Galois if it is normal and separable. But since $\pi_{\alpha,K}$ has at least one root in $L$, ($\alpha$ itself) then it contain all roots of $\pi_{\alpha,K}$. Then the extension is the large enough from your claim, hence 1) and 2) holds.
Now, since your $\pi_{\alpha,K}(x)= x^d+c_{d-1}x^{d-1}+...+c_0=(x-\alpha_1)...(x-\alpha_d)$ The identities 3 and 4 follow from the Viete’s formula (https://en.wikipedia.org/wiki/Symmetric_polynomial), as you wanted.