Answer
See work below.
Work Step by Step
If a photon is emitted, the electron moves to a lower state, so the final n is 1, 2, 3, 4, or 5.
We started in a d subshell, with $\mathcal{l}=2$. On page 814, the selection rule tells us that for allowed transitions, $\Delta \mathcal{l}=\pm 1$. Therefore, the final value of $\mathcal{l}$ must be 1 or 3.
a. A p subshell has $\mathcal{l}= 1$ and a f subshell has $\mathcal{l}=3$. The possible values of $\mathcal{l}$ are from 0 to (n-1), so the possible final states are 2p, 3p, 4p, 5p, 4f, 5f.
Expressed as ordered pairs of (n, $\mathcal{l}$) as requested, the possible final states are (2, 1), (3, 1), (4, 1), (5, 1), (4,3), (5,3).
Incidentally, we see that a 6d electron cannot drop directly into a n=1 state.
b. In a hydrogen atom, the energy depends largely on n. Thus, there are four different n-values (n = 2, 3, 4 or 5) among the possible final states, and four different wavelengths of photon emission.
