Hints:

1) For a 2 component system, you can always think of this as analogous to a spin 1/2 Hamiltonian. So, here , and

2) Griffiths Eq. 7.51 is quite general - you could change the value of "a" (the Bohr radius) and E1 (the ground state energy) to whatever you want, that's the beauty of writing F(x) instead of E(R). So, there's not much new work to do, here! You know the minimum of F(x) (it's in my notes, you checked it last week with MMA). Think back to Griffiths' problem 6.26 (couple weeks ago) to see how you change the Bohr radius in a "muonic hydrogen" atom.

3) Although it looks like a straightforward problem at first, there is a surprising amount of algebra in this problem. I combined my equations (eliminating c_b) to get a 2nd order ODE with constant coefficients for c_a, which I then remembered was pretty easy to solve. (You need to find the quadratic "characteristic" equation, which has two roots.) In the end, you should get a general linear combo of two familiar functions, with unknown coefficients. But then you can find c_b in terms of those SAME coefficients (remember, once you know c_a, you know it's derivative, and then you have an equation that TELLS you c_b!) , and finally your boundary condition c_a(0) =1, c_b(0)=0 is used to get rid of all the unknowns. (In my lecture notes I gave what I found for c_b(t), after some simplification.)

4) If you integrate a delta function Integrate[delta[t'-t0],{t',-big, +big}], i.e. your integration limits go PAST its argument, you always get one. If your integration limits don't span the argument, you always get zero. But what happens if you integrate RIGHT up to the argument, i.e. Integrate[delta[t'-t0], {t', -big, t0}]. What do you get? It seems ambiguous, but Griffith's problem 2.24 tells you the answer is 1/2. You will really need that for this problem! Just take the Differential equations and integrate them from 0 to time T. If T<t0, the answer is easy. If T>t0, the answer is easy, but requires that you know a not-yet-known cb(t0). So let T=t0 and use Griffiths' trick...