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).
Example b) is not even flat, so it is not etale.
Morally, what's happening at $T=0$ in example c) is that the fiber is changing from $d$ distinct points to one point: when $d>1$, this doesn't fit with the idea of etale maps as being like covering maps. (The technical term for this is that the map ramifies at $T=0$ - as etale is equivalent to flat + unramified, this provides one proof that the map in c) is not etale.)
Before performing the computations to show that condition 3) fails for example c), I should note that the condition as written in your post at the time of this answer is not correct. It should instead read as follows:
If $f:X\to Y$ is a morphism of schemes with $y=f(x)$ for some $x\in X, y\in Y$, then $\mathfrak{m}_y\mathcal{O}_{X,x} = \mathfrak{m}_x$ and the field extension $\kappa(y)\subset\kappa(x)$ is separable.
To check flatness, we observe that $\Bbb C[T,Y]/(Y^d-T)$ is a finitely-generated free $\Bbb C[T]$ module with basis $1,Y,Y^2,\cdots,Y^{d-1}$. By computing fiber products, we see that the fiber over $T=\lambda$ is just the spectrum of the ring $R=\Bbb C[T,Y]/(Y^d-T,T-\lambda)$. But $R\cong\Bbb C[Y]/(Y^d-\lambda)$, and when $\lambda\neq0$, $Y^d-\lambda$ splits as a product of linear terms, so $R\cong \Bbb C^d$ and it is immediate to check condition 3). When $\lambda=0$, the condition on maximal ideals fails: $\mathfrak{m}_{y=0}\mathcal{O}_{X,x=0}=(Y^d)$ which is not even maximal.
For example d), the map is etale precisely because the locus of points where 1) or 3) failed was removed.
Best Answer
If $X$ is a smooth connected affine curve whose completion has genus $> 0$, then $H^1(X_{cx}, \mathbb Z)$ will be non-zero. (It will have rank $\geq 2g$.) On the other hand $H^1_{et}(X,\mathbb Z)$ will vanish. (This latter claim is not immediately obvious, but true. As you would have learned from the links in your MO crosspost, it's not true in certain singular cases, e.g. if $X$ is a nodal curve.)
Basically the point is that in algebraic geometry we can see connected components, and finite covering spaces, but not irreducible, infinite degree, covering spaces. If you look at my posts in the SBS thread I linked to on your MO post, you will see that the non-vanishing $H^1(X,\mathbb Z)$ for a nodal curve is related to the fact that the resulting copy of $\mathbb Z$ is not really coming from an infinite degree irreducible cover, but is rather counting the components of a reducible cover (a chain of lines covering the nodal curve).