Just in case anyone is interested, this is what I've worked up so far:
There is a canonical isomorphism
(G/A)/(B/A) ≅ G/B
If you can construct a homomorphism φ:G/A → G/B, show that it is well-defined, then show that its kernel is B/A, you're done with all parts of the problem.
For B, this amounts to showing φ does not depend on choice of coset representatives. Assume you have x and y such that φ(xA) ≠ φ(yA). Prove that xA ≠ yA.
For C, just work straight from the definition of homomorphism. Specifically, set it up by choosing arbitrary x, y ∈ G, then show φ(xA yA) = φ(xA)φ(yA). It's pretty much immediate.
For D, suppose you have b ∈ B. Where does φ(bA) map? φ(bA) = bB = eB which is the identity, so bA ∈ ker φ. This proves A as well since the kernel of a homomorphism is always a normal subgroup.
The canonical isomorphism listed above is the conclusion of part E.