For part a), I have a question: if you're required to change to polar coordinates, then precisely what is in your original Lagrangian? Is it merely the x-coordinate, like you say? Or is it the vector ? Either would make sense in your original Lagrangian. If is merely the x-coordinate, then how come there isn't a in the original Lagrangian? It just seems a bit strange.

Normally, you wouldn't do that. The beauty of the Euler-Lagrange equations is that they apply to any set of generalized coordinates. Let's say I have a LagrangianBut i'm not sure of some things, like do i first solve for x in the extremal and use this as my general coordinate in the euler-lagrange equation?

and I transform it into polar coordinates:

The Euler-Lagrange equations before the transformation would have been

The Euler-Lagrange equations after the transformation are

So you see that the form does not change, even when you do a change of variables. What you want to do is apply this second system of equations to your transformed Lagrangian, without reference to back-solving for x or anything like that. Just stay in the new variable system.

For part b), the same thing applies. Hamilton's equations look the same for a coordinate and its conjugate momentum, regardless of what coordinates you choose. So do the transformation to whatever variables you want to use, find the conjugate momenta for each new coordinate, plug those momenta back into your Lagrangian to eliminate the coordinate velocities, form the sum that defines the Hamiltonian, and then write down Hamilton's equations. At least, that's the usual procedure. There might be some fancy shortcut somewhere to help you out. But this is what is normally done.

For part c), I'm afraid I can't help you much. I haven't studied the Hamilton-Jacobi equation in enough detail to really know what I'm talking about.