$\newcommand{\dcl}{\mathrm{dcl}} \newcommand{\acl}{\mathrm{acl}} \newcommand{\eq}{\mathrm{eq}} \newcommand{\tp}{\mathrm{tp}}$
You're exactly right: Chatzidakis's condition 1 is not equivalent to EI. Condition 2 is equivalent to WEI, but the equivalence is not obvious (at least it wasn't obvious to me).
I'll write DCLF for the property that for every definable set $D$, there is a smallest $\dcl$-closed set over which $D$ is definable. Similarly, I'll write ACLF for the property that for every definable set $D$, there is a smallest $\acl$-closed set over which $D$ is definable. I'll show below that Chatzidakis's conditions 1 and 2 are equivalent to DCLF and ACLF, respectively.
I've borrowed the name DCLF from the paper Weak forms of elimination of imaginaries by Casanovas and Farré. This paper is a wonderful reference for all the various properties related to EI. Casanovas and Farré discuss DCLF on p. 132 and give the theory of infinite sets as a counterexample to the equivalence with EI, just as you did. Proposition 2.10 clarifies the exact sense in which DCLF is weaker than EI: DCLF plus coding finite sets of pairwise interdefinable tuples is equivalent to EI (compare with the fact that WEI plus coding arbitrary finite sets of tuples is equivalent to EI). Finally, Casanovas and Farré point out on p. 128 that Poizat's original paper on EI, Une Théorie de Galois Imaginaire, mistakenly claimed that EI and DCLF were equivalent, and this error has propagated to some other sources - I would guess that this is the origin of the mistake in the Chatzidakis notes.
On the other hand, ACLF is well-known to be equivalent to WEI (and it is sometimes taken as the definition of WEI, as you did in your question). But the fact that DCLF is not equivalent to EI means that I would prefer not to take ACLF as the definition of WEI. The usual definition is: for every imaginary element $e\in M^{\eq}$, there is a real tuple $a\in M$ such that $e\in \dcl^{\eq}(a)$ and $a\in \acl^{\eq}(e)$. Or if you prefer to give an "intrinsic" definition not referring to imaginaries: for every definable set $D$, there is a finite set $F$ of tuples such that an automorphism fixes $D$ setwise if and only if it fixes $F$ setwise. Both of these definitions are natural generalizations of conditions that do correctly characterize EI (replace $\acl^{\eq}$ with $\dcl^{\eq}$ in the first, and replace the finite set $F$ with a single tuple in the second).
Ok, enough discussion, on to the proofs. In the following arguments, I'll have to bounce back and forth between $M$ and $M^{\eq}$, so just to be clear about notation: $\acl^{\eq}$ is algebraic closure computed in $M^{\eq}$. When $B\subseteq M^{\eq}$, I'll write $B\cap M$ for the "real part" of $B$. When $A\subseteq M$ is real, $\acl(A)$ is the algebraic closure of $A$ computed in $M$, and note that $\acl(A) = \acl^{\eq}(A)\cap M$. The same goes for $\dcl$ and $\dcl^{\eq}$.
First, it is clear that DCLF implies condition 1 and ACLF implies condition 2. If there is a smallest $\dcl/\acl$-closed set $C$ over which $D$ is definable, then for any two $\dcl/\acl$-closed sets $A$ and $B$ over which $D$ is definable, $C$ is contained in $A\cap B$, so $D$ is definable over the $\dcl/\acl$-closed set $A\cap B$.
Conversely, suppose $T$ satisfies condition 2. We'll show it satisfies ACLF. Let $D$ be a definable set, and let $e\in M^{\eq}$ be its code. Let $\varphi(x,a)$ be any formula defining $D$ with $a\in M$. Then $e\in \dcl^{\eq}(a)$. By extension for algebraic independence (see here, for example) in $M^{\eq}$, we can find $b\in M$ with $\tp(b/e) = \tp(a/e)$ such that $\acl^{\eq}(ae)\cap \acl^{\eq}(be) = \acl^{\eq}(e)$. Since $\tp(b/e) = \tp(a/e)$, $e\in \dcl^{\eq}(b)$, so $D$ is definable over $b$. Then by condition 2, $D$ is definable over the $\acl$-closed set $C = \acl(a)\cap \acl(b)$. If $D$ is definable over some other $\acl$-closed set $C'\subseteq M$, then $e\in \dcl^{\eq}(C')$, so $$C\subseteq (\acl^{\eq}(a)\cap \acl^{\eq}(b))\cap M \subseteq \acl^{\eq}(e)\cap M \subseteq \acl^{\eq}(C')\cap M = \acl(C') = C'.$$
Thus $C$ is the smallest $\acl$-closed set over which $D$ is definable.
Finally, suppose $T$ satisfies condition 1. We'll show it satisfies DCLF. Let $D$ be a definable set, and let $e\in M^{\eq}$ be its code. Since condition 1 is stronger than condition 2, the proof above shows that $D$ is definable over $\acl^{\eq}(e) \cap M$. For each finite tuple $a$ in $\acl^{\eq}(e) \cap M$, let $N_a$ be the number of realizations of $\tp(a/e)$. Pick some finite tuple $c\in \acl^{\eq}(e) \cap M$ such that $D$ is definable over $c$ and such that $N_c$ is minimal. I claim that $C = \dcl(c)$ is the smallest $\dcl$-closed set over which $D$ is definable.
First we'll show that $C$ is a minimal $\dcl$-closed set over which $D$ is definable. Let $C'\subseteq C$ be a $\dcl$-closed set such that $D$ is definable over $C'$. Let $c'\in C'$ be a finite tuple over which $D$ is definable. Since $c'\in C = \dcl(c)$, $N_{c'}\leq N_c$. But by minimality of $N_c$, $N_{c'} = N_c$. It follows that $c\in \dcl(c')$. But then $C = \dcl(c) \subseteq \dcl(c')\subseteq C'$.
Now let $B$ be any $\dcl$-closed set over which $D$ is definable, and let $b\in B$ be a finite tuple over which $D$ is definable. By condition 1, $D$ is definable over $\dcl(b)\cap \dcl(c)\subseteq B\cap C \subseteq C$. By minimality, $C = B\cap C$, so $C\subseteq B$. Thus $C$ is the smallest $\dcl$-closed set over which $D$ is definable.
You have to distinguish between what individual sentences can do and what theories can do. Theories, being arbitrary sets of sentences, are able to perform arbitrarily large conjunctions in the following sense:
For any set of theories $\{T_i: i\in I\}$, the theory $T:=\bigcup_{i\in I}T_i$ "behaves like the conjunction of the $T_i$s" in the following sense: the models of $T$ are exactly those structures satisfying each $T_i$.
On the other hand, even if each $T_i$ consists just of a single sentence $T_i=\{\varphi_i\}$, there may be no single sentence $\psi$ such that the models of $\psi$ are exactly the models of each $\varphi_i$. For a concrete example, there is no single sentence true in exactly the infinite structures, even though for each $n$ there is a sentence true in exactly the structures of size $\ge n$.
Best Answer
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 $\mathrm{dcl}(a,b)$.