When lightning strikes a car, it is not simply charge separation that protects the occupants; the conductive skin provides a preferential path for a current. If the charge were to simply be deposited on the skin of the car, something exciting might happen, but in fact the charge is carried through the car to the ground. It is definitely possible to apply a large enough current that ordinary conduction no longer occurs, but this isn't because the metal skin runs out of electron/hole pairs -- it's because the skin heats up and vaporizes!
Your premise for part 2 isn't quite correct. If you stick an electron inside a neutral, conductive sphere, you can still see the charge outside the sphere. If the sphere is grounded, then you will not, but that's because you've essentially hidden the charge on the ground and given the sphere an opposing charge.
Conductive shells shield their contents from external fields, but they do not shield the external environment from the charges of their contents.
If you add charge, grounding will continue to cancel your fields. If you increase the charge rapidly, then you can gain a momentary field, but currents will eventually cancel the charge accumulation. (Essentially, you are charging a capacitor, which is leaky because of connection to the ground.)
To return to the question you posed in part 1; what happens if we apply a slowly increasing field to a hollow conductive sphere that is perfectly insulated from ground? At first, polarization will shield the interior of the sphere from the external electric field. The conduction electrons will flow to create a polarization across the sphere. Eventually, however, the material will breakdown.
Exactly how is a matter of conjecture, on my part. Because the sphere will polarize, with electrons accumulating on one side and bare ion lattice on the other, I hypothesize that the ends of the sphere will begin to smear out perpendicular to the electric field. The electric field balances the repulsion between like species along the electric field, but not across it. Therefore, the ends of the sphere will experience a shear that thins them and breaks them. These broken pieces, being charged, then fly along the electric field at colossal speeds, allowing electric field into the erstwhile interior, and badly damaging your lab.
This statement shall provide you the answer.
"The electric field inside a conductor is 0 and all excess charge resides at the surface of the conductor."
So, no matter whether you have a solid or a hollow metal sphere(that is a conductor) there exists no electric field inside it and all the charge you give resides at the surface of the sphere.
Then what about the role of the dieletric? It turns out that it has no effect. The electric field inside the conductor will still be zero. This is because the electrons are free to move inside a conductor and they ultimately cause the field inside it to be zero.
This is very similar to the phenomenon of electrostatic shielding.
A small explanation of why is it so is given here
Best Answer
The important conditions here are that you are considering the interior of a conducting surface with the topology of a spherical shell, at equilibrium. Its spherical symmetry and surroundings turn out to be irrelevant. The spherical shell will be at an electric equipotential. (If it weren’t, the gradient of the potential would make charge redistribute.) The entire interior will have the same value of electric potential (because solutions of Laplace’s equation cannot have local maxima or minima). Any apertures in the shell would invalidate the conclusion.