(3) Making Donaldson's construction work equivariantly seems far from trivial. One can lift the cirle-action circle-action to the line bundle $L\to M$ whose curvature is the symplectic form. We now want a sequence of approximately holomorphic sections $(s_k^0,s_k^1)$ of $L^k$, satisfying estimated transversality conditions, such that $(s_k^0:s_k^1)$ is $S^1$-invariant. But we won't be able to make the sections themselves $S^1$-invariant, S^1$-equivariant, because the construction involves bombarding$M$with Gaussian sections; they live in little coordinate charts, while the circle-orbits may be big. 1 Hm, good question. This would be potentially interesting from the point of view of classifying Hamiltonian circle-actions, "by induction" on the dimension. (1) I'm pretty sure that the answer is not known to be "yes". A MathSciNet search of papers citing Donaldson's paper on symplectic Lefschetz pencils turns up nothing. (2) It sounds plausible to me (provided that one assumes that the symplectic class is integral). Of course, someone may be spot a counterexample. (3) Making Donaldson's construction work equivariantly seems far from trivial. One can lift the cirle-action to the line bundle$L\to M$whose curvature is the symplectic form. We now want a sequence of approximately holomorphic sections$(s_k^0,s_k^1)$of$L^k$, satisfying estimated transversality conditions, such that$(s_k^0:s_k^1)$is$S^1$-invariant. But we won't be able to make the sections themselves$S^1$-invariant, because the construction involves bombarding$M\$ with Gaussian sections; they live in little coordinate charts, while the circle-orbits may be big.