Question 1: I suppose it depends on what you mean by "interesting". There are certainly many examples in which the correspondence is meaningful, in the sense that for a typical definably-closed set $A$, there are many definably closed sets $C$ with $A\subseteq C \subseteq \mathrm{acl}(A)$.
Just staying within the realm of field theory, you could look at the theories of separably closed fields (SCF), pseudo-algebraically closed (PAC) fields, pseudofinite fields, differentially closed fields (DCF), algebraically closed fields with a generic automorphism (ACFA), etc. Outside of field-theoretic examples, you could consider finitely branching trees, or any theory with a definable equivalence relation with infinitely many finite classes, etc. Of course we have to avoid any theory with a definable linear order, since this implies that $\mathrm{acl}(A) = \mathrm{dcl}(A)$ for all sets $A$.
Question 2: A silly answer is to take the theory of an equivalence relation $E$ with infinitely many infinite classes and add all the imaginary sorts necessary to code finite sets. The resulting theory does not have codes for $E$-classes, so it codes finite sets but does not eliminate imaginaries.
Here's a more natural answer: ACVF (the theory of algebraically closed valued fields), or more generally any theory of fields that fails to eliminate imaginaries. Indeed, fields always code finite sets (using the trick with elementary multi-symmetric polynomials. But they need not eliminate imaginaries, e.g. if $K$ is a model of ACVF, the equivalence relation "same coset modulo the maximal ideal" on the valuation ring (whose equivalence classes are the elements of the residue field) is not eliminated.
Question 3: I'm not aware of any general criteria. Coding finite sets is a fairly concrete condition, so to show a theory $T$ codes finite sets, usually you just give the coding explicitly. On the other hand, if you want to show that $T$ does not code finite sets, a common strategy is to find an automorphism $\sigma$ of the monster model that swaps a pair $(a,b)$, i.e., $\sigma(a) = b$ and $\sigma(b) = a$, so that $\sigma$ fixes the unordered pair $\{a,b\}$, and show that $\sigma$ fails to fix any tuple in $\mathrm{dcl}(a,b)$.
Best Answer
Basically it comes down the treatment of non-connected covers / non-transitive actions. Classical Galois theory – whether applied to field extensions or to covering spaces – focuses on connected covers and transitive group actions. Every transitive left action of a (discrete) group $G$ is isomorphic to the set of left cosets of some subgroup of $G$ (with the obvious left $G$-action), which is how subgroups enter the picture.
One way to make this precise is to define connectedness abstractly.
Definition. A connected object in a category $\mathcal{C}$ with finitary (resp. infinitary) coproducts is an object $X$ in $\mathcal{C}$ such that $\mathcal{C} (X, -) : \mathcal{C} \to \textbf{Set}$ preserves finitary (resp. infinitary) coproducts.
Remark. According to this definition, an initial object is never connected.
Example. A $G$-set is connected in the sense above if and only if it is transitive.
Example. A ring $A$ is connected in $\textbf{CRing}^\textrm{op}$ considered as a category with finitary (!!!) products if and only if $\operatorname{Spec} A$ is a connected topological space. (In this context, $\emptyset$ is not connected.) Equivalently, $A$ is connected in $\textbf{CRing}^\textrm{op}$ if and only if $A$ has exactly two idempotent elements, namely $0$ and $1$.
Example. A finite étale algebra over a field $k$ is connected in $\textbf{FÉt}_k{}^\textrm{op}$ if and only if it is a finite separable field extension of $k$.
So we can extract the objects studied in classical Galois theory, at least. To get the actual posets is a bit more difficult. This should not be surprising: after all, in the context of field extensions, this amounts to choosing an algebraic closure and embedding all the field extensions into that algebraic closure. But it can be done: this is what the fibre functor is for.
Let $\mathcal{C}$ be a category and let $U : \mathcal{C} \to \textbf{Set}$ be a functor. We may form the following category $\textbf{El} (U)$:
An object is a pair $(X, x)$ where $X$ is an object in $\mathcal{C}$ and $x \in U (X)$.
A morphism $(X, x) \to (Y, y)$ is a morphism $f : X \to Y$ in $\mathcal{C}$ such that $U (f) (x) = y$.
Composition and identities are inherited from $\mathcal{C}$.
Incidentally, $U : \mathcal{C} \to \textbf{Set}$ is representable if and only if $\textbf{El} (U)$ has an initial object.
In the case where $\mathcal{C}$ is the category of connected $G$-sets and $U$ is the forgetful functor, $\textbf{El} (U)$ is a preorder category, which can be canonically identified with the poset of open subgroups of $G$: just send $(X, x)$ to the stabiliser subgroup of $x$.
In the case where $\mathcal{C}$ is the opposite of the category of finite separable field extensions of $k$ and $U$ is the functor sending $K$ to the set of $k$-embeddings $\iota : K \to \bar{k}$, where $\bar{k}$ is a chosen algebraic closure of $k$, $\textbf{El} (U)$ is a preorder category, which can be canonically identified with the opposite of the poset of finite subextensions of $\bar{k}$: just send $(K, \iota)$ to the image of $\iota : K \to \bar{k}$.
Since Grothendieck's formulation asserts that the opposite of the category of finite étale $k$-algebras is equivalent to the category of finite $\textrm{Gal} (k)$-sets as categories equipped with fibre functors, restricting to the subcategory of connected objects and applying the construction above recovers the classical antitone isomorphism of posets.