The answer to your question is a qualified no. Part of the reason is that we can't assume that every object in our model is named by an individual constant. So for instance, it could be that our model satisfies the sentence $\bigwedge_{i \in I} \neg P(c_i)$, where "$\bigwedge_{i \in I}$" indicates (possibly infinite) conjunction over index set $I$, and $c_i$ are all constants of the language, and yet this same model also satisfies the sentence $\exists x P(x)$. It's just that the object in our model which satisfies $P(x)$ is unnamed.
Of course, you're right that there is a strong analogy between quantifiers and infinite conjunctions/disjunctions in the following sense: if we require that every object in our domain is named by a constant, and if we allow for arbitrary conjuncts/disjuncts, then we can translate the quantified sentences into quantifier-free sentences using (possibly infinite) conjunctions/disjunctions. Logicians sometimes define substitutional quantifiers for this purpose: for instance, letting $\Sigma$ be a new substitutional quantifier, $\Sigma x \varphi(x)$ is true in a model just in case for some constant $c$, $\varphi(c)$ is true in that model, i.e. just in case $\bigvee_{i \in I} \varphi(c_i)$ is true in that model, where $I$ indexes the constants of $L$.
With that said, an infinitary propositional logic without quantifiers is not the same as a first-order logic with quantifiers. For one thing, in a propositional logic, you can only say $p$ is true or false. Your models aren't collections of objects with structure, but rather are simply truth value assignments for the proposition letters. So it's hard to say in what sense, if any, an infinitary propositional logic is the same as first-order logic without infinitary conjunctions/disjunctions. Their models don't even look alike.
Furthermore, even an infinitary predicate logic without quantifiers fails to be equivalent to first-order logic with quantifiers (but only finite conjunctions/disjunctions). The reason is simple: in first-order logic, there is no sentence which is true exactly when the domain is infinite. However, if the language you invoke has (at least) countably many constants $c_i$, then the sentence $\bigwedge_{\substack{i,j \in \omega \\ i \neq j}} c_i \neq c_j$ can only be true in infinite models. Hence, infinitary predicate logic without quantifiers is not compact, and so can't be equivalent to first-order logic.
We have always problem in formalizing natural language statements.
The first step is how to translate : whenever.
We assume that it has the same meaning of "when".
Thus, the statement is of the form :
When $A$, then $B$
and we symbolize it with the connective : $\rightarrow$ ("if ..., then _") :
$A \rightarrow B$.
Now we need quantifiers for analyzing the two clauses :
The first one will be :
$\exists x(Alert(x) \land Active(x))$
while for the second we have :
$\forall y((Message(y) \land Queued(y)) \rightarrow Transmitted(y))$.
Putting all together :
$$\exists x(Alert(x) \land Active(x)) \rightarrow \forall y((Message(y) \land Queued(y)) \rightarrow Transmitted(y))$$
Best Answer
Formulas (1) and (4) are valid, i.e. they are true in every first-order $\mathcal{L}$-structure.
Formulas (2) and (3) are not valid, i.e. there exists a $\mathcal{L}$-structure in which they are not true. For instance, take the $\mathcal{L}$-structure $\mathcal{N}$ whose domain is $\mathbb{N}$ and whose interpretation of $P$ is $2\mathbb{N}$ (the set of even natural numbers), and whose interpretation of $Q$ is $\mathbb{N} \smallsetminus 2\mathbb{N}$ (the set of odd natural numbers). You have that the formula $\forall xP(x) \Rightarrow \forall x Q(x)$ is vacuously true in $\mathcal{N}$ (it claims that "if every natural number is even then every natural number is odd"), but the formula $\forall x(P(x) \Rightarrow Q(x))$ is false in $\mathcal{N}$ (it claims that "for every natural number, if it is even then it is odd"), therefore your formula (2) is false in $\mathcal{N}$. Similarly for the formula (3), since $A \Leftrightarrow B$ is equivalent to $(A \Rightarrow B) \land (B \Rightarrow A)$.
In general, when one talks about distributivity of something over something else (for instance, distributivity of $\land$ over $\lor$), one means that two formulas are logically equivalent. With this meaning, the answer to your question "Does the universal quantifier distribute over conditional or biconditional?" is negative since the formula $\forall xP(x) \Rightarrow \forall x Q(x)$ is not logically equivalent to the formula $\forall x(P(x) \Rightarrow Q(x))$ (your formula (1) is valid, but your formula (2) is not valid), and similarly the formula $\forall xP(x) \Leftrightarrow \forall x Q(x)$ is not logically equivalent to formula $\forall x(P(x) \Leftrightarrow Q(x))$ (your formula (4) is valid, but your formula (3) is not valid).