= Solution
Let $R=\prod_i\mathcal M_i/F$ be the <reduced product> by a proper <filter on a set>. Its underlying <equivalence relation> is $f\sim g$ when $\{i:f(i)=g(i)\}\in F$. Operations are interpreted coordinatewise, and a relation holds of the classes exactly when its coordinate truth set belongs to $F$.
Evaluation of a <first-order term> commutes with passage to the quotient, by <mathematical induction> on terms. Consequently the desired equivalence holds for every <atomic formula>, including <logical equality>. For a formula $\theta$ and representatives $\bar f$, write $S_\theta=\{i:\mathcal M_i\models\theta(\bar f(i))\}$.
For <logical conjunction>, $S_{\theta\wedge\psi}=S_\theta\cap S_\psi$. The <filter on a set> axioms give
$$
S_\theta\cap S_\psi\in F\quad\Longleftrightarrow\quad S_\theta\in F\text{ and }S_\psi\in F.
$$
Thus the induction hypothesis transfers a conjunction in both directions.
For <existential quantification>, first suppose $R\models\exists y\,\theta(y,[\bar f])$. Choose a representative $g$ of a witness. Induction gives $S_{\theta(g,\bar f)}\in F$, and this set is contained in $S_{\exists y\theta}$. Upward closure therefore gives $S_{\exists y\theta}\in F$.
Conversely, suppose $A=S_{\exists y\theta}\in F$. For each $i\in A$, choose a coordinate witness $g(i)$, and choose an arbitrary element of $M_i$ outside $A$. These simultaneous choices use the <axiom of choice>, as does the usual product construction. Then $A\subseteq S_{\theta(g,\bar f)}$, so that truth set belongs to $F$. Induction gives $R\models\theta([g],[\bar f])$, providing the required witness. Therefore
$$
\boxed{R\models\varphi([\bar f])\iff\{i:\mathcal M_i\models\varphi(\bar f(i))\}\in F}
$$
for every <primitive positive formula>. The exam's <tame formulas> are exactly this fragment, built using <logical conjunction> and <existential quantification>. No <ultrafilter> dichotomy was used.
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