How does Galois group acts on etale cohomology? I know this may be a trivial question, but I can't find the answer on, for example, Milne's online notes and Danilov's Cohomology of Algebraic Varieties.
Suppose $K$ is a number field (say), $\overline K$ its algebraic closure, $G_K:=\text{Gal}(\overline K/K)$, $X$ is a variety over $K$, $\mathcal F$ is a locally constant (or constructible (?), don't know if it's true) abelian sheaf on $X_{\text{et}}$. It is said that for any $i\geqslant 0$, there is a continuous $G_K$ action on $H_{\text{et}}^i(X_{\overline K},\mathcal F)$.
My question is, how can one define this action? I found the answer that if $i=1$ and $\mathcal F$ is the constant sheaf associated to an abelian group $A$, then $H_{\text{et}}^1(X_{\overline K},\mathcal F)=\text{Hom}(\pi_1^{\text{et}}(X_{\overline K}),A)$,
using the following exact sequence
$$
0\to\pi_1^{\text{et}}(X_{\overline K})\to
\pi_1^{\text{et}}(X)\to\pi_1^{\text{et}}(\text{Spec }K)\cong G_K\to 0
$$
we get the Galois action. But what about $i\neq 1$ or $\mathcal F$ is not constant sheaf?

[EDIT] After viewing other questions on this site, I found that it is the proper base change theorem: if $f:X\to\text{Spec }K$ is proper, then there is a canonical isomorphism $H_\text{et}^i(X_{\overline K},\mathcal F_{\overline K})\xrightarrow\sim(R^if_*\mathcal F)_{\overline K}$, the latter is a $G_K$-module. (Complaint: this theorem is contained in notes I mentioned, but it relates etale cohomology to Galois representations is unmentioned, strange.)
But what about $X$ not proper, for example $X=Y_1(N)$ the open modular curve?

[EDIT2] Thanks for Roland's answer, but could anyone explain the simplest example: when $X=\text{Spec }k$, why does this definition of Galois action on $H_{\text{et}}^0$ coincides with the action given by the following category equivalence?
\begin{align*}
\mathbf{Sh}\big((\text{Spec }k)_{\text{et}},\mathbf{Ab}\big)
&\cong\{\text{discrete abelian group with continuous }G_k\text{-action}\} \\
\mathcal F&\mapsto\varinjlim_{k'/k\text{ finite separable extension}}
\mathcal F(\text{Spec }k') \\
\big(\text{Spec }k'\mapsto A^{\text{Gal}(\overline k/k')}\big)&\leftarrow A
\end{align*}
In fact, at first I thought I understand this category equivalence, but later I found that I didn't.
 A: Please correct me if I am wrong, but I really don't see where we need the proper assumption to get a Galois action.
So let $X/k$ be any scheme, $\overline{k}$ a separable (or algebraic) closure of $k$, and $\mathcal{F}$ be any sheaf on $X$. Write $X_\overline{k}$ for the base change of $X$, $p:X_\overline{k}\rightarrow X$ the induced morphism. Write also $\mathcal{F}_\overline{k}=p^*\mathcal{F}$ for the pullback of $\mathcal{F}$ to $X_\overline{k}$.
For $g\in\mathrm{Gal}(\overline{k}/k)$, we have an induce morphism $g:X_\overline{k}\rightarrow X_\overline{k}$, which induce a pullback $g^*\mathcal{F}_\overline{k}$. I claim that there is a canonical isomorphism $g^*\mathcal{F}_\overline{k}\simeq\mathcal{F}_\overline{k}$. This is because $\mathcal{F}$ comes from $X$. Indeed : $g^*\mathcal{F}_\overline{k}=g^*p^*\mathcal{F}=p^*\mathcal{F}=\mathcal{F}_\overline{k}$.
Hence we have an induced action on cohomology :
$$ H^i(X_\overline{k},\mathcal{F}_\overline{k})\rightarrow H^i(X_\overline{k},g^*\mathcal{F}_\overline{k})\simeq H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
which is the action you are looking for.
Now there is also the approach of Alex Youcis : if $f:X\rightarrow\operatorname{Spec}k$ is the structural morphism, $R^if_*\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$ hence a set equipped with a continuous discrete $\operatorname{Gal}(\overline{k}/k)$-set. Unless I'm mistaken, I don't think we need to add any assumption for the following claim : the underlying $\operatorname{Gal}(\overline{k}/k)$-set is exactly $H^i(X_\overline{k},\mathcal{F}_\overline{k})$ with the above action. (This prove in particular that the above action is continuous).
This follows from the following continuity result 
$$\varinjlim_{k'}H^i(X_{k'},\mathcal{F}_{k'})=H^i(X_\overline{k},\mathcal{F}_\overline{k})$$
where $k'$ runs through the finite Galois extensions of $k$ in $\overline{k}$ and the trivial base change $(R^if_*\mathcal{F})_{k'}=R^if_*\mathcal{F}_{k'}$.

EDIT : Let me expand a bit on the last assertion. First, let us show that the stalk of $R^nf_*\mathcal{F}$ at a geometric point $\operatorname{Spec}\overline{k}$ is indeed $H^n(X_\overline{k},\mathcal{F}_\overline{k})$.
By definition, the stalk of $R^nf_*\mathcal{F}$ is $\varinjlim_{k'}R^nf_*\mathcal{F}(k')$ where the limit is taken over all the finite extension of $k'$ inside $\overline{k}$. Now recall that $R^nf_*\mathcal{F}$ is the sheaf associated to the presheaf $k'\mapsto H^n(X_{k'},\mathcal{F}_{k'})$, and since the stalk of a presheaf is the same as its associated sheaf, we get $(R^nf_*\mathcal{F})_\overline{k}=\varinjlim_{k'}H^n(X_{k'},\mathcal{F}_{k'})=H^n(X_\overline{k},\mathcal{F}_\overline{k})$, the last equality is from a limit argument (we should add $X$ quasi-compact and quasi-separated here).
Recall that the equivalence between sheaves on $(\operatorname{Spec}k)_{ét}$ and $\operatorname{Gal}(\overline{k}/k)$-sets is the following : if $\mathcal{F}$ is a sheaf on $\operatorname{Spec}k$, then $\mathcal{F}_\overline{k}=\varinjlim\mathcal{F}(k')$ is a $\operatorname{Gal}(\overline{k}/k)$-set. The action of $\sigma\in\operatorname{Gal}(\overline{k}/k)=\varprojlim\operatorname{Gal}(k'/k)$ on $\mathcal{F}_\overline{k}$ is induced by the compatibles actions on the $\mathcal{F}(k')$ (check that is indeed compatible).
Now you should convinced yourself that this is indeed the pull-back to $\overline{k}$ and that this action is the same as the induced action by functoriality (that is very first one I wrote). Do the same with $R^if_*$ and you will get the compatibility of the two actions in general.
Exercise : Take $k=\mathbb{R}$ and $\mathcal{F}=\mu_4$ the sheaf $A\mapsto\{x\in A, x^4=1\}$ and compute the action of the stalk (there is no trap here).
