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"?
Let's consider for simplicity the spherically symmetric case, where we have a spherical cavity and a point particle in the middle. In the presence of non-zero charge density we need to consider the Poisson equation, $\nabla^2 \phi = - 4\pi \rho$. Here $\rho(x) = q \delta(x)$ is the charge density of the point particle which is placed at $x=0$ (the middle of the cavity). Notice that there is a singularity at $x=0$, so the situation is not the same as with an empty cavity: we cannot assume that the solution is smooth at $x=0$. This is why a constant potential is not a solution in this case.
One way to treat the singularity is to 'cut out' a small ball $B$ of radius $r$ around the point charge from the space, and then to treat the surface of this ball as another boundary (in addition to the boundary that is the conducting shell). The new space exists between the outer conducting and the small sphere that surrounds the charge. This way, the solution will be smooth everywhere in the new space. But we now need to specify the boundary conditions on this inner surface, which should somehow be related to the charge we removed. Let's work out these boundary conditions. Integrating the Laplace equation over the ball $B$, we have
\begin{align}
\int_B \nabla^2 \phi = -4\pi \int_B \rho = -4\pi q \,.
\end{align}
Using Stoke's theorem, the integral on the left becomes
\begin{align}
\int_{\partial B} ds \, \hat{n} \cdot \vec{\nabla} \phi = 4\pi r^2 \partial_r \phi \,.
\end{align}
Here $\partial B$ is the boundary of the ball, namely the small shell of radius $r$. $\hat{n}=\hat{r}$ is the normal vector, and $ds$ is an area element on the shell.
We find the boundary condition
\begin{align}
\partial_r \phi(r) = - \frac{q}{r^2} \,.
\end{align}
Therefore, the potential should have a non-vanishing gradient close to the point charge (as expected), and this rules out a constant solution in the cavity. Of course, we know the solution in this case is simply $\phi(r) = q/r$, which is consistent with the boundary conditions on the conductor and close to the point charge.
Best Answer
A conducting body can have a potential, and it need not be zero. Potential can be arbitrarily set, depending upon your reference potential. The only difference in the tratement of conducting bodies is that they must be equipotential, i.e., they must have constant potential at all points inside them (but not necessarily points inside cavities).
The potential of a metal shell due to its own charge $q_1$ is $\frac{kq_1}{r_{shell}}$ If you add a point charge $q_2$ at the center, then the potential becomes $\frac{kq_1}{r_{shell}} + \frac{kq_2}{r_{shell}}$.
Remember, potential at a point
iscan be defined with respect to the work required to get a test charge there from infinity. If the field at a point is zero, that doesn't imply that the field is zero all the way to infinity. It just means that you can jiggle a test charge in the neighborhood of that point without doing work.