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"?
Same charges repulse each other. So when they are confined in a system, they try to have stable distance among them as much as possible. For a hollow conducting sphere this stable maximum distance is equal distribution of charges in the outer surface. If any charge try to go the inside conducting surface it automatically decrease distance which increase repulsion. That's why charge inside the conducting surface zero.
Best Answer
The statement "electric field inside a conductor is zero" is true only after charges have distributed themselves in the most optimal way on the surface - it is an electrostatic result. Starting with an arbitrary charge distribution, there will be forces that cause a redistribution of the charge until, for a sphere, they are distributed uniformly. At that time, there is no electric field inside the conductor, and so no force on the charges that impels them to move to another, energetically more favorable, location.
A simple proof for spherical conductor is this: if the sphere is symmetrical, then the solution must also be symmetrical (there is nothing about a sphere that would drive an asymmetrical solution, and the uniqueness theorem says that if you have "a" solution that meets the boundary conditions, it must be "the" solution. Since uniform distribution meets the boundary conditions, it must be the solution.). But if that is so, then the electric field inside the sphere must also be spherically symmetrical. And we know from Gauss's theorem that the integral $\int E\cdot dS$ must equal the $\frac{Q}{\epsilon_0}$. Since $Q=0$, it follows that $E=0$.