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"?
Well, your statement is incomplete. Electrostatic field is zero inside the MATERIAL of the conductor. It may have any other value inside( inside does not include the material of the conductor) or outside the conductor.
It is zero inside the MATERIAL of the conductor, because since charges are free to move, they generate a electric field due to polarised conductor, which tends to cancel out the external electric field.
With reference to your question, if in case 1 , the field was $E_{1}\vec{i}$, then in case 2 it would remain the same because electric field lines of external field and the polarised charges of the outer surface would not reach the inside of the conductor(as can be seen from diagram depicting field lines). Thus the RESULTANT( not the contribution due to individual), of external and outer surface charge would cancel each other.
Now if we take a cylindrical gaussian surface just enough to enclose the inner surface charges, since we know that field inside MATERIAL of conductor is zero, thus application of gauss law gives net charge enclosed as zero, which further yields that inner surface must have uniform distribution of $-q$ charge( uniform only because the filament is AT CENTER, if it would have been off center, distribution would be non uniform), thus by invoking symmetry arguements(or gauss law, if you want) we get that field due inner surface charge is also zero. Therefore filed will be only due to filament, thus case 2 is equivalent to case 1.
Best Answer
No, it does not imply that the surface is spherical and charged uniformly.
Imagine a charged conducting shell of arbitrary shape. (An ellipsoid is a simple example.) Gauss' Law tells us that the charges in the conductor fly to the outside surface of the conductor, and the distribution of charges is such that the E-field inside is zero. But for a non-spherical conductor, the charge distribution is explicitly not spherical, and the charge distribution on it is not uniform.