Reducible 3d torus bundles Here reducible means that the mapping class for the fiber is a reducible auto-homeomorph in the sense of Nielsen-Thruston. So,
could anyone give me a hint to classify them?
In contrast, do you agree that -in the sense of connected sum- all these bundles are irreducible?
 A: OK, time to give some references for this classical material. Orientable 3-manifolds that are torus bundles are classified up to diffeomorphism by the conjugacy class of their monodromy map in SL(2,Z). [More precisely: two monodromy matrices A and B in SL(2,Z) determine diffeomorphic torus bundles if and only if A is conjugate in GL(2,Z) to either B or the inverse of B.] If the monodromy is elliptic (finite order) or parabolic (having an eigenvalue 1 or -1) then the torus bundle is Seifert fibered, so the classification in these cases is covered by the general classification of Seifert manifolds. If the monodromy is hyperbolic (two distinct real eigenvalues) then the torus bundle has a solvgeometry structure. One possible reference for the more topological parts of this classification is my notes on 3-manifolds, available on my webpage, where Section 2.2 gives the topological classification of orientable torus bundles in some detail, including the conjugacy classification of elements of SL(2,Z). For the "eight geometries" aspects of the classification there is the well-known expository article of Peter Scott, "The Geometries of 3-Manifolds", in the 1983 Bull.L.M.S. (a scanned pdf version is available on his webpage). See pp. 470-472 in particular.
The case of nonorientable torus bundles should be similar. Scott actually treats the nonorientable case, but I was lazy and didn't go that far. There are probably other references too that I'm forgetting at the moment. Can anyone else remember?
In particular, to answer the original question, torus bundles are irreducible both in the Thurston geometries sense and in the embedded spheres sense.
A: Yes - surface bundles over the circle are irreducible (*) as long as the fiber is not a two-sphere.  This follows from the fact that the universal cover of such a surface bundle is homeomorphic to $\mathbb{R}^3$ and Proposition 1.6 of Hatcher's three-manifold notes. 
(*) in the sense of connect sum.
EDIT: To answer the question posed by Juan in the comment. A orientation preserving homeomorphism $h$ of $T^2$ is reducible (that is, preserves the isotopy class of some essential multicurve) if and only if $h$ is a power of a Dehn twist or is the power of a Dehn twist followed by the hyperelliptic element.  Here is a "cut-and-paste" proof: if $h$ preserves a multicurve then it preserves a curve, say $\alpha$.  Now, $h$ either preserves or reverses the orientation of $\alpha$.  If the latter case replace $h$ by $h$ followed by the hyperelliptic, to reduce to the former case.  Isotope $h$ so that $\alpha$ is fixed pointwise.  Note that, as $h$ preserves orientation of $T^2$, the sides of $\alpha$ are preserved as well.  Thus $h$ restricts to a homeomorphism of the annulus, fixing the boundary pointwise.  By the classification of mapping classes in the annulus, $h$ is a Dehn twist.
As a bit of an advertisement: Farb and Margalit have written a primer on the maping class group.  You can find a discussion of the mapping class groups of the disk, annulus, and torus in Section 2.4, on the "Alexander Method".  (In particular they give the usual proof that the group of orientation oreserving classes on $T^2$ is $SL(2,Z)$.  They also give as an exercise the characterization of Dehn twists.)
I'll end by pointing out that there is not a contradiction between my answer and Hatcher's.  If the monodromy is reducible then the resulting torus bundle is Seifert fibered and in fact has Nil geometry. 
A: I think the hint you need is that the (orientation-preserving) mapping class group of the torus is SL_2(Z).  So you need to figure out which elements are "reducible", and what the corresponding bundles look like.
