Answer
$Q_1=1.53\times10^{-3}C$
$Q_2=4.16\times10^{-3}C$
$V=611V$
Work Step by Step
Equations needed:
$Q=CV$ (pg 483)
$Q_1=C_1V_1=(2.50\mu F)(746V)=1870\mu C$
$Q_2=C_2V_2=(6.8\mu F)(562V)=3820\mu C$
$\sum Q=1870\mu C+3820\mu C=5690\mu C$
Once the two capacitors are connected, there will be no voltage difference, it will all reach equilibrium, so $V_1=V_2=V_f$
$Q_1+Q_2=5690\mu C$
$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$
$C_2Q_1=C_1Q_2$
$C_2 5690\mu C-C_2Q_2=C_1Q_2$
$(6.8\mu F)(5690\mu C)-(6.8\mu F)Q_2=(2.5\mu F)Q_2$
$Q_2=4.16\times10^{-3}C$
$Q_1=1.53\times10^{-3}C$
$V=\frac{1.53\times10^{-3}C}{2.5\mu F}=611V$