Suppose $\mathbf{B}$ is a complete Boolean algebra with an infinite domain $B$. Suppose $\mathbf{B}$ is atomic (i.e. every element is the supremum of some set of atoms). This algebra contains the co-finite filter $\mathscr{F}_c$, which in case the set of atoms of $\mathbf{B}$ is countable is generated by a chain. This follows from the fact that we may arrange all atoms in a sequence $a_0,a_1,\ldots$ and take: $$C:=\{B-\{a_0,\ldots,a_n\}\mid n\in\omega\}$$ as a chain that generates $\mathscr{F}_c$. It is known that no free ultrafilter in $\mathbf{B}$ is generated by a chain (see this answer).

In light of the above my first question is:

(1) If $\mathbf{B}$ has a countable set of atoms, can it contain a free filter which is different from $\mathscr{F}_c$ and which is generated by a chain?

Considering an uncountable case:

(2) If $\mathbf{B}$ is atomic with uncountably many atoms, is the co-finite filter $\mathscr{F}_c$ generated by a chain?

(3) If $\mathbf{B}$ is atomic with uncountably many atoms, can it contain a free filter which is different from $\mathscr{F}_c$ and which is generated by a chain?