Given, (A) n=5, ~m_ 1 =-1 (B) n=3, l=2, ~m_ 1 =-1, ~m_ s =+(1)/(2)...

For case (A) with n=5, $m_l=-1$:
The principal quantum number n=5 allows l from 0 to 4. Since $m_l=-1$, we need l≥1. For each l value (1, 2, 3, 4), the orbital $m_l=-1$ 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, $m_l=-1$, $m_s=+\frac{1}{2}$:
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.