I don't have definitive answers to your questions, but I'll try to make some comments.
For 1), I haven't done a literature search to see if this was known. However, Boyer and Clay propose a related open question Problem 1.11.Problem 1.11.
For 2), as I mentioned in the comments above, any irreducible manifold $M$ containing an incompressible torus admits a Reebless foliation. This follows from Theorem 5.1 of Gabai's paperTheorem 5.1 of Gabai's paper. Cut $M$ along an incompressible torus to get manifolds $M_1, M_2$. Then let these be taut sutured manifolds with boundary being $R_+$. By Theorem 5.1, there is a foliation containing $\partial M_i$ as leaves, and which is taut in the interior (hence Reebless). Then glue these two foliations together to get a Reebless foliation of $M$. A caveat here is that the foliation might not be smooth.
Hence, double branched covers of prime alternating knots which have a Conway sphere admit Reebless foliations but not taut ones. The preimage of a Conway sphere is an incompressible torus, but by Ozsvath-Szabo, the double branched cover is an L-space, so does not admit a (smoothe co-oriented) taut foliation.
Simpler examples might come from manifolds admitting taut non-oriented foliations, but no taut oriented foliation, for example sol manifolds that semi-fiber (with trivial first betti number).
For 3), I would suggest that the answer is yes, in the sense that before the work of Ozsvath and Szabo, we knew of very few classes of manifolds not admitting taut oriented foliations (hence, it appears that for most of their examples, Heegaard Floer homology cannot be bypassed). We knew that there was an algorithmthere was an algorithm to (in principle) detect if a given manifold admits a taut foliation, but it is not practical to implement. Something akin to this was implemented to find infinite collections of manifolds admitting no taut foliationcollections of manifolds admitting no taut foliation (see also Calegari-DunfieldCalegari-Dunfield). The big open question though is whether the converse might hold: if (irreducible orientable) $M$ does not admit a taut (smooth co-oriented) foliation, then is it an L-space? I think Juhasz may have posed this as a question or conjecture.