This is a little bit tricky to see straight away because you've got some overlapping peaks.
A: NH - I agree
B: chloroform (from the NMR solvent).
That leaves C, D and E.
You've called E a quartet, and assigned this as the methyl groups 8 and 11. This signal is for Me groups 8 and 11, but it is not a quartet, it is two separate signals. Me 8 and 11 are chemically inequivalent, and so will give two separate signals. Each of them couples to an adjacent CH2, so we would expect to see two separate 1:2:1 triplets for methyl groups 8 and 11. Now, in this case these two signals are very close together and overlap slightly.
I will try to explain, you have two triplets: 1:2:1 1:2:1
But they overlap like this: 1:2:1
1:2:1
So it looks like a 1:3:3:1 quartet, but in reality is not.
The overlapping is nothing other than bad luck.
Hint: Consider this kind of situation when looking at signal C
What splitting patterns would you expect from the H at positions 7, 10, 12 and 13?
Hint: For signal D, you have measured a coupling constant of 21 Hz, which implies you think it's a doublet. 21 Hz is too high for a H-H coupling constant (0-15 is the usual range)...