For case (A) with n=5, ml=−1:
The principal quantum number n=5 allows l from 0 to 4. Since ml=−1, we need l≥1. For each l value (1, 2, 3, 4), the orbital ml=−1 exists, giving 4 orbitals. Each orbital holds 2 electrons (spin up and down), so maximum electrons = 8.
For case (B) with n=3, l=2, ml=−1, ms=+21:
These quantum numbers completely specify a single electron in a unique orbital with a specific spin, so maximum electrons = 1.
Therefore, the answer is 8 and 1.