2 retracted over-optimistic idea

This isn't an answer but a suggestion of how to get a consistent answer. The idea is to assume the continuum hypothesis (CH) and build a Hausdorff gap with an additional property.

By a Hausdorff gap, I mean two $\omega_1$-sequences $(A_\alpha)$ and $(B_\alpha)$ of infinite subsets of $\omega$, each almost increasing (here and below, "almost" means "modulo finite sets"), with each $A_\alpha\cap B_\beta$ finite, and such that no set $C$ almost includes every $A_\alpha$ and is almost disjoint from every $B_\alpha$. [Caution: The same name, "Hausdorff gap" is used with different meanings by lots of people. Some replace the $B_\alpha$'s by their complements. Some impose additional requirements, making the gap resemble in certain ways the ones constructed by Hausdorff. The latter people don't all agree on what resemblance to require.] Hausdorff's construction works in ZFC and is somewhat tricky. If one assumes CH, there's a brute force construction: List all the subsets $C$ of $\omega$ in an $\omega_1$-sequence and construct the gap by induction, defining $A_\alpha$ and $B_\alpha$ at step $\alpha$ in such a way as to ensure that the $\alpha$-th $C$ in the list doesn't violate the definition of Hausdorff gap.

Consider (for the moment) any Hausdorff gap; let $I$ and $J$ be the ideals generated by the set of $A_\alpha$'s and the set of $B_\alpha$'s, respectively. Let $F$ and $G$ be the filters obtained from $I$ and $J$ as in the question. Then $F\cup G$ generates a proper filter. [Otherwise, there would be some set $C\in F$ with $\omega-C\in G$, and that would violate the definition of Hausdorff gap.] For this filter to be an ultrafilter, you need that, whenever $\omega$ is partitioned into two pieces, then one of the pieces, say $P$, has the property that $(A_\alpha\cap P)$ and $(B_\alpha\cap P)$ do not form a Hausdorff gap, i.e., there should be a set $C$ that almost includes every $A_\alpha\cap P$ and is almost disjoint from every $B_\alpha\cap P$. An arbitrary Hausdorff gap is unlikely to satisfy this. Nevertheless, it seems to me that this requirement can be met by listing all the relevant partitions in an $\omega_1$-sequence and, at each stage $\alpha$ in the construction of the gap, ensuring that one of the pieces of the $\alpha$-th partition has the property we want.

If this works, is shows that, under CH, the answer to Question 1 is positive. I would expect that, if it works, then a modification of the argument would work under Martin's axiom (or just $\mathfrak p=\mathfrak c$). [EDIT: I retract the preceding sentence. I envisioned a $\mathfrak c$ long construction of a $(\mathfrak c,\mathfrak c)$ gap, but such a construction runs the risk of ending prematurely with, for example, a Hausdorff gap.] But it doesn't look likely to me that an argument like this can be made to work in ZFC without any special hypotheses. Even though one can construct Hausdorff gaps in ZFC, I don't see any way to "mix" that $\omega_1$ long construction with handling all $\mathfrak c$ partitions if $\mathfrak c>\aleph_1$. Notice, though, that the argument doesn't need the full strength of a Hausdorff gap; it doesn't matter that the sequences $(A_\alpha)$ and $(B_\alpha)$ are almost increasing, so we could work with something like Lusin gaps instead. Unfortunately, I don't see how that extra freedom helps.

1

This isn't an answer but a suggestion of how to get a consistent answer. The idea is to assume the continuum hypothesis (CH) and build a Hausdorff gap with an additional property.

By a Hausdorff gap, I mean two $\omega_1$-sequences $(A_\alpha)$ and $(B_\alpha)$ of infinite subsets of $\omega$, each almost increasing (here and below, "almost" means "modulo finite sets"), with each $A_\alpha\cap B_\beta$ finite, and such that no set $C$ almost includes every $A_\alpha$ and is almost disjoint from every $B_\alpha$. [Caution: The same name, "Hausdorff gap" is used with different meanings by lots of people. Some replace the $B_\alpha$'s by their complements. Some impose additional requirements, making the gap resemble in certain ways the ones constructed by Hausdorff. The latter people don't all agree on what resemblance to require.] Hausdorff's construction works in ZFC and is somewhat tricky. If one assumes CH, there's a brute force construction: List all the subsets $C$ of $\omega$ in an $\omega_1$-sequence and construct the gap by induction, defining $A_\alpha$ and $B_\alpha$ at step $\alpha$ in such a way as to ensure that the $\alpha$-th $C$ in the list doesn't violate the definition of Hausdorff gap.

Consider (for the moment) any Hausdorff gap; let $I$ and $J$ be the ideals generated by the set of $A_\alpha$'s and the set of $B_\alpha$'s, respectively. Let $F$ and $G$ be the filters obtained from $I$ and $J$ as in the question. Then $F\cup G$ generates a proper filter. [Otherwise, there would be some set $C\in F$ with $\omega-C\in G$, and that would violate the definition of Hausdorff gap.] For this filter to be an ultrafilter, you need that, whenever $\omega$ is partitioned into two pieces, then one of the pieces, say $P$, has the property that $(A_\alpha\cap P)$ and $(B_\alpha\cap P)$ do not form a Hausdorff gap, i.e., there should be a set $C$ that almost includes every $A_\alpha\cap P$ and is almost disjoint from every $B_\alpha\cap P$. An arbitrary Hausdorff gap is unlikely to satisfy this. Nevertheless, it seems to me that this requirement can be met by listing all the relevant partitions in an $\omega_1$-sequence and, at each stage $\alpha$ in the construction of the gap, ensuring that one of the pieces of the $\alpha$-th partition has the property we want.

If this works, is shows that, under CH, the answer to Question 1 is positive. I would expect that, if it works, then a modification of the argument would work under Martin's axiom (or just $\mathfrak p=\mathfrak c$). But it doesn't look likely to me that an argument like this can be made to work in ZFC without any special hypotheses. Even though one can construct Hausdorff gaps in ZFC, I don't see any way to "mix" that $\omega_1$ long construction with handling all $\mathfrak c$ partitions if $\mathfrak c>\aleph_1$. Notice, though, that the argument doesn't need the full strength of a Hausdorff gap; it doesn't matter that the sequences $(A_\alpha)$ and $(B_\alpha)$ are almost increasing, so we could work with something like Lusin gaps instead. Unfortunately, I don't see how that extra freedom helps.