Answer
(a) The potential difference does not change.
(b) The capacitance decreases by a factor of $\frac{1}{2}$
(c) The charge decreases by a factor of $\frac{1}{2}$
Work Step by Step
(a) Since the capacitor is still connected to the battery, the potential difference does not change. The potential difference is still $\Delta V_{bat}$
(b) We can write an expression for the capacitance:
$C = \frac{\epsilon_0~A}{d}$
If the separation distance doubles, then the capacitance decreases by a factor of $\frac{1}{2}$
(c) We can write an expression for the charge:
$Q = C~\Delta V$
If the potential difference remains the same while the capacitance decreases by a factor of $\frac{1}{2}$, then the charge decreases by a factor of $\frac{1}{2}$
