Let $F$ be a sheaf of Abelian Groups on the Zariski site. Let us extend the definition as described in the question of the Zariski sheaf $F$ to a functor defined on the etale site and call it $F_{et}$. We first verify that this is indeed a sheaf.
First certain preliminaries.
Let $g : U \rightarrow X$ be an etale open set in $X$, then $F_{et}(g:U \rightarrow X) = F(g(U))$. This is easy to see by the definition of the pullback of a sheaf and the fact that $g(U) \subset X$ is an open subset in the zariski topology, since $g$ is etale in particular flat. Next, let us see what happens on the intersection of two such open sets.
Let $g : U \rightarrow X$ and $h : V \rightarrow X$ be two open subsets. Let $g\times_X h : U \times_X V \rightarrow X$ be the fiber product. Then $F_{et}(g \times_X h : U \times_X V \rightarrow X) = F((g \times_X h)(U \times_X V \rightarrow X)) = F(g(U) \cap h(V))$. The first equality follows as above while the second equality follows from the fact that $(g \times_X h)(U \times_X V \rightarrow X) = g(U) \cap h(V)$. This is explained below :
We have the following setup $U \rightarrow g(U) \hookrightarrow X$, where the first arrow is surjective and the second is an open immersion(topologically and if we give (and we do) $g(U)$ the induced subscheme structure, then scheme theoretically). A similar factorization exists for the arrow $h$. Using the universal property of cartesian diagrams we get that $g \times_X h$ factors through $g(U) \times_X h(U) = g(U) \cap h(V)$(This is clear since $g(U)$ and $h(V)$ are subsets of $X$). Let us analyze the map $U \times_X V \rightarrow g(U) \times_X h(U)$.
Claim : The above map $U \times_X V \rightarrow g(U) \times_X h(U)$ is surjective.
Proof of Claim : Note that this map factors as follows $U \times_X V \rightarrow g(U)\times_X V \rightarrow g(U) \times_X h(U)$. Now, note that $U \rightarrow g(U)$ and $V \rightarrow h(V)$ is surjective. We know that base change of a surjective morphism is surjective. Hence both the arrows are surjective maps and hence the composition is surjective. Thus proving the claim.
Now, we prove that $F_{et}$ is infact a sheaf. Let $\lbrace g_i : U_i \rightarrow U \rbrace$ be an etale cover of an etale open set $g : U \rightarrow X$ i.e. each of the $g_i$ is an etale map and $\cup_i g_i(U_i) = g(U)$. We consider the standard sequence of Abelian groups
$F_{et}(g:U \rightarrow X) \rightarrow \prod^i F_{et}(g_i) \rightarrow \prod F_{et}(g_i \times_X g_j)$
From above paragraphs, the exact sequence is same as the exact sequence
$F(g(U)) \rightarrow \prod F(g_i(U_i)) \rightarrow F(g_i(U_i) \cap g_j(U_j))$
This is an exact sequence follows from the fact that $F$ is a sheaf on the zariski site.
Thus we have that $F_{et}$ is infact a sheaf. Note that the etale cohomology of this sheaf will be the same as usual cohomology as can be seen by considering cech complexes.
This is infact functorial for the morphism for sheaves.
Regarding the first question, I've looked at the example Alex Youcis mentioned, that is $X=\mathrm{Spec}(B)$, $B=\prod_{\mathbb{N}}\mathbb{F}_2$ which is homeomorphic to $\{0,1\}^\mathbb{N}$ ; on that space we have a topology that is totally disconnected (see Qiaochu Yuan's blog post on boolean rings). Then we take $A=\mathbb{F}_2$ and compute :
$A_X(X)=\mathrm{Hom}_X(X,X\coprod X)=\mathrm{Hom}_B(B\times B,B)$.
Now a B-linear map $\varphi \colon B\times B \to B$ is characterized by the image of $e_1=(1,0)$ and $e_2=(0,1)$ (since as an $B$-module, $B\times B=B\oplus B$), which we denote by $f_1$ and $f_2$, satisying certain conditions so that it is a ring map : $f_1 + f_2 = 1$, $f_1f_2=0$. This means exactly that $f_1$ are $f_2$ are a pair of associated idempotents, and so choosing an idempotent in $B$ determines the pair. Now since every element of $B=\prod_{\mathbb{N}}\mathbb{F}_2$ is idempotent, we have a bijection $A_X(X))\simeq \{0,1\}^{\mathbb{N}}$.
On the other hand, since $X$ is totally disconnected,
$\prod_{\text{connected components of $X$}}\mathbb{F}_2 \simeq \mathbb{F}_2^{\mathrm{card}(X)}=\mathbb{F}_2^{\{0,1\}^{\mathbf{N}}}\simeq \{0,1\}^{\{0,1\}^{\mathbf{N}}}$
which means the LHS is in bijection with the powerset of $A_X(X)$, so it can't be isomorphic to the latter.
As for the second question, I'll cite the answer given by Alex Youcis in the comments for completness :
This is an old chestnut of a question. The answer is that you're right--the second equality doesn't a priori hold for arbitrary $X$. Note though that if $X$ is Noetherian since any étale map $Y \to X$ is locally of finite presentation you know that $Y$ is locally Noetherian. It's certainly not Noetherian in general since something like $\coprod_i \mathrm{Spec}(L_i)\to\mathrm{Spec}(K)$ is étale where the index set can be arbitrary. But, as this example shows, note that if $Y \to X$ is étale then we can get an étale cover of Y of the form $\coprod_i U_i\to Y$ where $U_i$ are Noetherian opens inside of $Y$. Then, on $\coprod_i U_i$ the desired equality does hold since each of the $U_i$ are Noetherian. So, if $X$ is Noetherian you have a cofinal system of étale covers for which the desired equality does hold, which is usually good enough in practice.
Edit : Actually if $X$ is only locally Noetherian, any étale $X$-scheme is also locally Noetherian, hence has open connected components since it is locally connected. Indeed, pick an étale $X$-scheme $Y$, $Y_0$ a connected component of $Y$ and $y\in Y_0$. Then $y$ has an affine open noetherian neighbourhood $U$, and $U$ is the finite union of its connected components, which must thus be open, and $y$ must belong to one of them, say $V$. Then $V$ has to be a subset of $Y_0$ and is open in $Y$.
Best Answer
You are right: $X_{ét}$ is large (in fact, essentially large). This means that there is no category of presheaves on $X_{ét}$ in ZFC, or if you are using universes you would need to go to a higher universe to talk about the category of presheaves.
However, the category of sheaves $Sh(X_{ét})$ is indeed a genuine category that can be defined without enlarging your universe. This is because there is a small set of objects of $X_{ét}$ which can be used to cover all other objects, and so a sheaf is uniquely determined by the values it takes on a small subcategory of $X_{ét}$. Indeed, since an étale map is locally of finite presentation, it suffices to consider affine schemes which are finitely presented étale covers of affine open subsets of $X$, and there is an (essentially) small set of these.