(Ah, a very old question which is not yet answered... and I don't find a duplicate answer, either, though the problem is very nice ...
the comment is point you there, and that the user "centralcharge" has edited the question is amazingly fitting, too :) but it's not yet elaborated)
Yes, you can use the method of images. Normally you are right, it requires the object to be grounded, but for a sphere there is a nice "trick".
You have to first find an image, that would make the potential on the sphere zero. How to do this is apparently clear to you. This would solve the grounded case.
But there is no requirement for the potential to be zero. It has just to be equal on the sphere. And this requirement is not violated, if you put some charge in the center of the sphere. This central charge will just shift the potential on the sphere, but obviously not break spherical symmetry.
Take the extra charge just as big as you need to adjust the right net charge. Then you have solved the problem for non-grounded sphere.
So now to your full problem: In what distance would the sphere with $+Q$ net charge and the point charge of $+Q$ be at equillibrium?
You can solve this with an quite complicated equation or just guess the solution and check. Let's guess the golden ratio: $r = \frac{\sqrt 5 + 1}{2}R$, and briefly check this:
I'll set the radius of the sphere and the charges to $1$. Now assume the outside charge has distance $\phi = \frac{\sqrt 5 + 1}{2}$ from the center of the sphere. Then the induced charge is $-1/\phi$ and it's distance from the center is $1/\phi$ too. The charge in the middle has to be $1+1/\phi$ for the net charge to be $1$
Now it holds $1/\phi = \phi-1$, that's why $\phi$ is nice. It follows:
The force between the outside charge and the induced charge: the distance is $1$, the product of charges is $1/\phi$, so the force is an attraction of $\frac{1/\phi}{1^2} = 1/\phi$.
The force between the outside charge and the central charge: the distance is $\phi$, the product of charges is $\phi$, so the force is a repulsion of $\frac\phi{\phi^2} = 1/\phi$.
The forces are equal, qed.
(...if you solve the equations you see, that this is indeed the only real solution greater than 1)
Question 1:
Yes. If the PEC (perfect electric conductor) is grounded it has a fixed potential of zero volts on all points of the boundary. That's a standard Dirichlet boundary conditions in differential equation terminology. The charge $Q$ on the conductor will generally not be zero, but will be given by the Gauss equation:
$$\oint_\Gamma\vec{E}\cdot\vec{\mathrm{d}A}=\frac{Q}{\varepsilon_0}$$
In the other scenario, you do not know the potential on the boundary, and IMHO you can not call it a Dirichlet condition. PEC's are still equipotential, you just don't know what this potential is. Instead, you are given the charge $Q$ (zero in the case of neutrality) and the Gauss equation must be solved as part of the problem to obtain the potential on the boundary.
It's worth noting that since $\vec E = -\nabla\phi$ any constant offset in $\phi$ is irrelevant. It's the electric field that eventually leads to motion of particles in the physical world (through the Lorentz' force), so in the end only relative potentials matter. This means that you can usually set the potential to zero at an arbitrary point of your choice, i.e. you get to choose the ground/reference node.
The insulator is irrelevant. It only changes the surrounding field, not the boundary conditions of the PEC.
Question 2:
Yes. It's the electric field that must be zero within a PEC, which according to $\vec E=-\nabla\phi$ means that the potential must be constant, not necessarily zero. Any practical circuit will have conductors at different voltages. Again, you get to choose ground, but the relative potentials must remain.
I can't quite make sense of how you would involve the image charges in this but I think the answer to that is: no, image charges do not necessarily make the potential non-zero. Since you get to choose the ground you could choose the object under consideration to be ground. In that case it would be zero, but it could still have image charges according to Gauss equation.
As an end-note: For non-ideal conductors you may have a slight electric field within it, and therefore a varying potential, but I take it that we're discussing ideal conductors here. Most metals have a conductivity in the order of $10^7\,\mathrm{S/m}$ so it's usually a valid approximation.
Question 3:
Without having seen the entire problem myself I assume Jackson decomposed the problem into two simpler problems.
First problem: It's likely simpler to solve the problem with the external charges by fixing the potential of the sphere to for instance zero volts, since this means you have Dirichlet boundary condition. However, by doing so the sphere will have a charge $Q'$ following from the Gauss equation for the determined solution and not what was specified ($Q$).
Second problem: The difference between the original problem and the "first subproblem" is that of a sphere with charge $Q-Q'$ with no external charges. We know that the fields produced by a sphere and point charge of equal charge in the middle of it is the same (this is only valid outside the sphere). This problem has a well known solution.
By linearity of the Poisson equation, the two solutions can be added to give the solution of the original problem (superposition).
Best Answer
If the sphere is initially uncharged, then the electric flux through its surface is zero, by Gauss' law. If we add just one point charge $q'$, then we will find a net flux through the surface, which is wrong. However, add a second charge $q^{''}=-q'$, then the net flux is given by the enclosed charge $q'+q^{''}=0$, and all is well.