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.
That's a very perceptive question.
When an insulator becomes polarized, all that has to happen is that for each molecule, there is a slight shift of electrons relative to the molecule. The point is that there are a LOT of electrons in any material - as you know, the charge of one electron is $1.6\cdot 10^{-19}$ C, and there are $N_A$ molecules per mole of material. So for an insulator like quartz (molecular weight 60), 60 grams of material would have electrons with a total of $(14+8+8)\cdot N_A\cdot 1.6\cdot 10^{-19}$ C of charge - 2.8 million Coulomb! That charge needs to be displaced by just a tiny amount to give you a significant polarization (we measure dipole moment as charge times displacement). The key here is that in an insulator, all the molecules play a role in the total polarization ... a tiny little contribution by each of a very large number of molecules.
By contrast, if you put a conductor in an electric field, a few electrons will move around on the surface and cancel the electric field inside. For those few electrons, the displacement will be quite large (compared to the movement of electrons in the insulator). Because there is no electric field inside the conductor, we say the material is not polarized (that is, there is no displacement between the electrons of a given atom, and the nucleus - because there is no field).
Best Answer
What the insulator does in this problem is keeping the shells electrically neutral. So that any positive charge built on the inner wall will be balanced with the negative charge on the outer wall. And your figure (unlabeled) of replacing the incubator layers with conductors is correct.