Partial pressure of gases in a container with a piston
$B=\pu{0.6226 bar}$ as the water-water vapour system would be in equilibrium.
Now,
$A+B= \pu{1 bar}\\
A= \pu{0.3774 bar}$
For initial volume of the container,
$
P_{\ce{Ar}_i}V_i=n_{\ce{Ar}_i}RT\\
V_i=C=\frac{n_{\ce{Ar}_i}RT}{P_{\ce{Ar}_i}}\\
C=\pu{\frac{0.1 \times 0.08314 \times 360}{0.3774} L}\\
C=\pu{7.93 L}\\
$
For initial moles of water vapour,
$
P_{\ce{H2O_{(v)}}_i}V_i=n_{\ce{H2O_{(v)}_i}}RT\\
n_{\ce{H2O_{(v)}_i}}=\frac{P_{\ce{H2O_{(v)}_i}}V_i}{RT}\\
n_{\ce{H2O_{(v)}_i}}=\pu{\frac{0.6226 \times 7.93}{360 \times 0.08314} L}\\
n_{\ce{H2O_{(v)}_i}}=\pu{0.165 mol}\\
$
Now,
$
n_{\ce{H2O_{(v)}_i}} + n_{\ce{H2O_{(l)}_i}} =\pu{1 mol}\\
n_{\ce{H2O_{(l)}_i}} = D = \pu{ (1 - n_{\ce{H2O_{(v)}_i}}) mol}\\
D = \pu{ (1 - 0.165) mol}\\
D = \pu{ 9.835 mol}\\
$
Now as the volume is doubled by Boyle's law we can say that the partial pressure of argon and water vapour will decrease to half of original.
Thus,
$
E=\frac{A}{2}=\pu{0.1887 bar}\\
F=\frac{B}{2}=\pu{0.3113 bar}\\
$
Now as the partial pressure of water vapour is less than the vapour pressure of water at that temperature, water will evaporate to make it equal.
Now as the temperature is same thus the vapour pressure of water is also same as initial. Thus only change in system is it's volume which was doubled. So we can conclude that twice amount of water vapour as compared to initial conditions will be required to attain this pressure.
Thus,
$
n_{\ce{H2O_{(v)}_f}}=2 \times n_{\ce{H2O_{(v)}_f}}\\
n_{\ce{H2O_{(v)}_f}}=\pu{0.33 mol}\\
$
But as the total moles of water is conserved,
$
G=H=\pu{(1-n_{\ce{H2O_{(v)}_f}}) mol}\\
G=H=\pu{0.67 mol}\\
$