First of all, potential for $r > b$ will not be constant.
You could solve the above problem by superposition.
* Potential due to inner conducting sphere only is $V_0$ for $r < a$ and $V_0*a/r$ for $r > a$
* Potential due to outer dielectric is $\sigma_0b^3\cos(\theta)/3\epsilon_0r^2$ for $r > b$ and $\sigma_0r\cos(\theta)/3\epsilon_0$ for $r<b$ (This can be calculated by treating the charge distribution as superposition of positively and negatively charged spheres. Ref. The Feynman Lectures on Physics Chapter 6)
Potential at $r=a$ is $V_0 + \sigma_0a\cos(\theta)/3\epsilon_0$
Now, as potential of conducting sphere must be constant, charges redistribute to give a potential of $-\sigma_0a\cos(\theta)/3\epsilon_0$ at $r=a$. We have seen that $\sigma = \sigma_0\cos(\theta)$ gives potential distribution of that kind.
Thus charge distribution on conducting sphere may be taken as $4\pi\epsilon_0aV_0 - \sigma_0\cos(\theta)$
Thus Potentials come out to be
$V_0$ for $r<a$
$V_0a/r+(r^3-a^3)\sigma/3r^2\epsilon_0$ for $b>r>a$
$V_0a/r+(b^3-a^3)\sigma/3r^2\epsilon_0$ for $r>b$
how can positive charges move to the far side? How can they move at all? I thought that positive metal ions don't move but only the electrons move (and since it's a conductor, I am assuming it consists of positively metal ions and free electrons).
The positively charged atomic nuclei do NOT move from one side of the sphere to the other. What happens is that some of the negatively charged electrons leave that side, leaving a net positive charge behind.
In electrolyte solutions and plasmas, positive ions do move, but in solids, the nuclei are essentially fixed in their position relative to their neighbors.
Regarding the same picture and charge distribution, I was also wondering if it is correct that a solid metal sphere can never have a volume charge, but only a surface charge, since the electric field must be zero inside a conductor always right?
That is essentially correct. A minor correction is that if there is a current in a conductor, there is an electric field $\vec{E}$ which is proportional to the current density $\vec{J}$ times the volume electrical resistivity of the conductor $\rho$.
$$\rho\vec{J}=\vec{E}$$
This is known as the "microscopic" version of Ohm's law.
Since the resistivity of most metals is quite small, the electric field will be small unless the current density is very large. The small electric field due a flow of current in a conductor is often ignored in practice.
However, even though there may be a small electric field within the body of a conductor in which there is an electric current, it is still true that charges do not accumulate in the interior of a conductor with uniform current density and uniform resistivity.
This can be shown by taking the divergence of the microscopic Ohm's law. If $\rho$ and $\vec{J}$ are uniform, then $\nabla \cdot \rho \vec{J} = 0$. Therefore $\nabla \cdot \vec{E} = 0$. Therefore by Gauss's Law, the charge density in that area of uniform current density $\vec{J}$ and uniform resistivity $\rho$ must also be $0$.
[Note that in this answer, I used $\rho$ as the symbol for resistivity. In the context of Gauss's Law, the symbol $\rho$ is often used for a different purpose, i.e. as the symbol for charge density. Please don't be confused by this.]
Best Answer
Short answer: yes, the surface charges are taken into account; in fact, they're what ensures that $\vec{E} = 0$ inside the conductor.
The electric field at any point in space can be viewed as the superposition of the fields from the point charge outside the sphere, and the induced surface charges: $$ \vec{E} = \vec{E}_\text{point} + \vec{E}_\text{induced} $$ Now, inside the conductor, the electric field must be zero; the usual argument for this is that if the electric field wasn't zero inside the conductor, then the charges would move around in response to it and we wouldn't have a stable configuration. So as we bring the point charge in from infinity towards the conducting sphere, the positive and negative charges rearrange themselves to cancel out the field inside the conductor. In other words, for points inside the conductor, we must always have $$ \vec{E}_\text{induced} = - \vec{E}_\text{point}. $$ The potentials inside the sphere, too, must cancel out to within a constant (namely, the potential of the sphere: $$ V_\text{induced} = \frac{q}{4 \pi \epsilon_0 x} - V_\text{point}. $$
A cute side-effect of this phenomenon (credit to Bob Geroch for posing a similar problem to me years ago) is the following: Suppose we could somehow freeze the induced surface charge in place on the sphere, and then remove the point charge. The electric field inside the sphere would then look exactly like there was a negative point charge at the same location outside the sphere, like an "electric afterimage". The equipotentials inside the sphere would be concentric arcs, centered at a point outside the sphere:
(Apologies for the clunky field line diagram; Mathematica is not well-adapted to making field line diagrams. The field lines do not, of course, end anywhere except at the surface of the sphere.)
For points outside the sphere, of course, this cancellation of the electric fields doesn't take place, and the electric field is non-zero. However, it is still the case that the potential is constant over the outer surface of the sphere; it must be so, or the electric field wouldn't vanish inside the conductor. If the sphere was an insulator, then points on the side of the sphere facing the charge would be at higher potential, and points on the side of the sphere away from the charge would be at lower potential. From the above diagram, it's not too hard to see that the effect of the surface charges is to lower the potential at points on the sphere that would otherwise be at higher potential, and vice versa; the net effect is that the sphere is at constant potential, as desired.