Is infinitary logic a logic in the sense of Gurevich?
First question: A set $M$ is decidable if there is a Turing Machine which halts on all inputs and accepts all inputs $x$ with $x \in M$.

We try to encode $\bigwedge_{\phi \in X} \phi$ for arbitrary sets of $\mathsf{FO}[\tau]$-formulars $X$. Since, $\mathcal{P}(X)$ is uncountable there can be no code with finite alphabet. Hence, there can be no Turing Machine recognizing $\mathcal{L}_{\infty\omega}[\tau]$ for each $\tau$ (even for empty $\tau$). Hence, the syntax is undecidable.

Second question: Given a code which translates a Deterministic Turing Machine (DTM) $T$ into a $\tau$-Structure $\mathcal{T}$. We can write a $\mathsf{FO}$-formular $\varphi_k$ with

$$ \mathcal{T} \models \varphi_k \Leftrightarrow \textrm{ The DTM $T$ holds after $k$ steps on $\varepsilon$.}$$

If we reassign variables (forgot the previous states of the Turing Machine) we can modell this $\varphi_k$ with constant, finite number $m$ of variables which does not depend on $k$.

Hence,

$$ T \in H_\varepsilon \Leftrightarrow \mathcal{T} \models \bigvee_{k \in \omega} \varphi_k \in \mathcal{L}_{\infty \omega}^m.$$

Where $H_\varepsilon$ is the special Halting Problem. Consequently, the Model Checking Problem for $\mathcal{L}_{\infty \omega}^m$ is undecidable for sufficiently large $m$.