Answer
$q_1 = 500~\mu C$
Work Step by Step
In part (a), we found that $q_3 = 750~\mu C$
Since $C_1$ and $C_2$ are in parallel, the total charge on capacitor 1 and capacitor 2 is $750~\mu C$. Note that $q_2 = q_3-q_1$
Since $C_1$ and $C_2$ are in parallel, the potential difference across capacitor 1 and capacitor 2 is equal.
We can find $q_1$:
$V_1 = V_2$
$\frac{q_1}{C_1} = \frac{q_2}{C_2}$
$q_1~C_2 = C_1~q_2$
$q_1~C_2 = C_1~(q_3-q_1)$
$q_1~(C_1+C_2) = C_1~q_3$
$q_1 = \frac{C_1}{C_1+C_2}~q_3$
$q_1 = (\frac{10.0~\mu F}{15.0~\mu F})~(750~\mu C)$
$q_1 = 500~\mu C$
