This is an old question, but it deserves a reasonable answer.
For Question 1:
Assuming that $F$ has genus $\ge 2$, isomorphism classes of $F$-bundles over any (smooth) manifold $M$ (e.g. the circle) are in a natural bijective correspondence with conjugacy classed of homomorphisms $\pi_1(M)\to MCG(F)$.
See Proposition 4.6 in S.Morita, "Geometry of Characteristic Classes". When $F$ has genus $\le 1$ the situation is a bit different, see the discussion in Morita's book preceding Proposition 4.6.
This answers your question 1 in the case of a (connected) manifold base. When the base is not a manifold, but is, say, a connected finite-dimensional CW complex $B$ one can prove the same thing by replacing $B$ with a (noncompact) smooth manifold homotopy equivalent to $B$. (I am not sure if you really want to see a proof since the question is quite old.)
As for Question 2, I am not sure how much do you want to know here. If total spaces of two fibrations are homeomorphic, the homeomorphism need not preserve the fibrations and things become quite complicated very quickly.
For Question 3: It depends on what do you mean by "cohomology". If you work with, say, rational, coefficients (or ${\mathbb Z}/p$ coefficients, $p$ is sufficiently large), then indeed $ H^*(MCG(F), {\mathbb Q})\cong H^*({\mathcal M}_g, {\mathbb Q}) $ where $g$ is the genus of $F$ and ${\mathcal M}_g$ is the moduli space of genus $g$ surfaces. This is because ${\mathcal M}_g$, treated as a DM stack (aka an orbifold), has contractible universal cover and the fundamental group isomorphic to $MCG(F)$. See page 146 of Morita's book. This is a special case of a more general theorem about non-free properly discontinuous group actions on CW complexes, you should be able to find this in Brown's book "Cohomology of groups". But if you work, say, with integer coefficients, then things are more complicated.