Free objects in first order theories
Let me explain why the question, raised in the comments, of what the morphisms of $M$ are, sheds a lot of light on this question. If $T$ is a first-order theory I will use $\mathrm{Mod}(T)$ to denote its category of models with elementary embeddings as morphisms, and $\mathrm{Hom}(T)$ to denote its category of models with homomorphisms as morphisms.
Observation: Let $M$ be a category and $U: M \to \mathsf{Set}$ a functor, and suppose that $U$ factors through the non-full subcategory $\mathsf{Set}_\mathrm{inj}$ of sets and injective maps. Suppose also that $U$ has a left adjoint. Then:
- $U$ must actually factor through the subcategory $\{\emptyset,1\}$.
- If moreover $U$ is faithful, then $M$ is a poset.
A proof will be given at the end. In particular, the hypotheses of both bullet points are satisfied in the following cases:
$M= \mathrm{Mod}(T)$ and $U$ the usual forgetful functor.
$M$ is $\mathsf{Set}_\mathrm{inj}$.
$(M,U)$ is any AEC, as in (Q3).
This leads to the following conclusions:
By (1) if in Q1 we take $M$ to mean $\mathrm{Mod}(T)$, then the answer is "$F$ exists only in degenerate cases".
By (2), the answer to Q2 is: "No".
By (3), the answer to Q3 is: "If $(M,U)$ is an AEC, then $F$ exists only in degenerate cases."
Let us also observe that
For any first-order theory $T$, there is a first-order theory $T'$ (the Morleyization of $T$) such that $\mathrm{Hom}(T') = \mathrm{Mod}(T)$, with the same forgetful functor to $\mathsf{Set}$.
$\mathrm{Hom}(T)$ need not be accessible. For example, let $T$ be the theory $\exists x \exists y x \neq y$. Then $M$ is the category $\mathsf{Set}_{\geq 2}$ of sets of cardinality at least 2, which doesn't even have split idempotents.
If $T$ is a complete theory with infinite models, then $\mathrm{Mod}(T)$ is never finitely accessible.
Now, if $M$ means $\mathrm{Hom}(T)$, we can conclude that
- On account of (1) the answer to Q1 is "There are lots cases where $F$ does not exist".
- On account of (2) and OP's observation that if $F$ exists then $M$ is finitely accessible, the answer to Q1 is "There is at least one case, and likely lots of cases, where $F$ does not exist".
- (1),and (3), together again with OP's observation about finite accessibility, give yet another way to see the answer to Q1 is "there are lots of cases for which $F$ does not exist.
But Q1 is pretty vague. It's possible that what the OP really means is "Even if it's rare for $F$ to exist, can we identify any conditions more general than cocompleteness under which $F$ exists?". In that case, the answer seems to be: "In light of the above observations, and especially (2), probably one can say no more about this question for $M = \mathrm{Hom}(T)$ than one can say about the case where $M$ is an arbitrary concrete category (which is to say, not much)".
Proof of Observation:
In $\mathsf{Set}$, every morphism except for the morphisms $\emptyset \to S$ where $S \neq \emptyset$ factors as a split mono followed by a split epi. But in $M$, every split mono and every split epi is an isomorphism. So the restriction $F|_{\mathsf{Set}_{\neq \emptyset}}: \mathsf{Set}_{\neq \emptyset} \to M$ of $F$ to the category of nonempty sets factors through the localization of $\mathsf{Set}_{\neq \emptyset}$ at all morphisms, which is the terminal category.
Now the coproduct $2 = 1 \amalg 1$ is preserved by $F$. The two coproduct injections $1^\to_\to 2$ are exchanged by an automorphism of 2 which is sent to the identity by $F$, so the two coproduct injections are the same map $F(1) \to F(2)$. This implies that for any object $X \in M$, there is at most one map $F(1) \to X$. But by adjointness, maps $F(1) \to X$ correspond to elements of $X$. So every object of $M$ has at most one element.
Moreover, $F(\emptyset)$ must be an initial object. This, along with the universal property of $F(1)$, implies the first bullet point. If $U$ is faithful, this means that $M$ must be a poset, with $F(\emptyset)$ a bottom element $\bot$. By its universal property, $F(1) = \min(U^{-1}(1))$. Moreover no element of $U^{-1}(1)$ is below an element of $U^{-1}(0)$. Conversely, any poset and functor $U$ meeting this description have a left adjoint $F$ meeting this description.