Answer
a) $1.5\;\rm V$
b) $ 2.12\times 10^{-11}\;\rm C$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The potential of an ordinary AA or AAA battery is
$$V=\color{red}{\bf 1.5}\;\rm V$$
$$\color{blue}{\bf [b]}$$
Since both, the battery and the capacitor, are connected in parallel, the battery will charge the capacitor until the potential difference between the capacitor plates is equal to the voltage of the battery.
Hence,
$$\Delta V_{\rm capacitor}=\color{blue}{\bf 1.5}\;\rm V$$
The charge in one plate is equal in magnitude to the charge on the other plate. So we can find the charge in one plate to know how much charge the battery supplied to each plate.
We know that
$$ V_C=Es$$
where the electric field inside the capacitor is $\eta/\epsilon_0$
$$ V_C=\dfrac{\eta s}{\epsilon_0}$$
where $\eta=Q/A$
$$ V_C=\dfrac{Qs}{A\epsilon_0}$$
Solving for $Q$,
$$ Q =\dfrac{A\epsilon_0V_C}{ s}$$
where the voltage from the 0 V plate (the negative one) to the V_C plate (the positive plate) is $s=d$
$$ Q =\dfrac{A\epsilon_0V_C}{ d}$$
Plug the known;
$$ Q =\dfrac{(4\times 4\times 10^{-4})(8.85\times 10^{-12})(1.5) }{ (1\times 10^{-3})}$$
$$Q=\color{red}{\bf 2.12\times 10^{-11}}\;\rm C$$
