 Solution

Membership of either truth set in a filter on a set implies membership of its union, by upward closure. The converse need not hold. For the cofinite filter, take $p(x)$ to mean that $x$ is even and $q(x)$ that $x$ is odd. The union of their truth sets is all of $\mathbb{N}$, while neither truth set is cofinite.

Thus the left side of the printed equivalence is true and its right side false: (ii) can be false. An ultrafilter does satisfy the equivalence, because its dichotomy forces one member of a finite union into the ultrafilter. General filters need not have that dichotomy. This is one aspect of the Boolean failure of the cofinite-filter quantifier.