You have two questions. One is "what is the charge distribution on a non symmetric conductor" and the other is "how can the inner surface of a conductor be charged?"
Q1:
The charges will distribute themselves in a manner to minimize total energy (whenever you don't know an answer on the exam, say that ;). Since it's a conductor, they will distribute themselves in such a way that the potential anywhere on the surface is the same (otherwise, the charges would have a gradient they could move along).
Q2
Assumptions:
Let's assume electro statics i.e. no time dependence, and perfect conductors.
Also let's assume the tubes are infinitely long. Therefore, this becomes a 2D problem of concentric circles (senior electromagnetis becomes much simpler when you realize they only ever test you on a handful of geometries).
Solution:
Point 1) The electric field inside a perfect conductor is zero (otherwise a current would arise, rearranging the charge until the field is cancelled out). Therefore, within the material of the thin outer tube, the electric field is zero.
Point 2) However, the integral of the electric field along a path (it's a path integral because we're in 2D!) is equal to the enclosed charge (Gauss' law). From symmetry, there is no angular dependence.
Conclusion) The only way for the first point and the second point to be both correct is if you have a charge in the inner surface of the tube. The charge on the inner surface of the tube needs to cancel out the electric field from the rod, so it is equal and opposite to the charge of the rod. The outer surface also has a charge, equal to the net charge of the tube minus the charge on the inner surface of the tube.
Remember:
The fundamental rule of a conductor is that the electric field within it is zero. There's no fundamental rule against charges on the inside surface of a conductor!
Further investigation:
Realistically charges are electrons in thermal equilibrium. How does this affect our mental image of "charges on an infinitely thin surface"?
The field of the induced charges is not used up to cancel the field in the metal of the inner charges, so that it could no longer be used for anything else.
The two effects (field of the charges inside the conductor and the external field) are independent. Each one accumulates charges on the outside of the conductor to kill the field inside of it. The effects are additive, so applying an additional external field simply will displace more charges to the surface of the conductor.
Mathematically speaking, you find the reason for why the effects are additive in the linearity of Maxwells equations. This is also the reason why the principle of superposition works.
Best Answer
The electric field everywhere in the conductor must be zero. This means that the electric field is zero everywhere on the wall of the cavity - and that in turn implies there can be no field in the cavity.
It follows that the distribution of charge on the surface is independent of the shape of internal cavities: if you consider a "skin" of conductor (thin layer) right outside the cavity, then the electric field at that point can be considered a boundary condition of the problem "where is the charge". And that boundary condition is unchanged whether the cavity is filled with more conductor, or with nothing.