(a) The increment in the internal energy is
$$
\Delta U=\int_{V_{1}}^{v_{1}}\left(\frac{\partial U}{\partial V}\right)_{T} d V
$$
But from second law
$$
\left(\frac{\partial U}{\partial V}\right)_{T}=T\left(\frac{\partial S}{\partial V}\right)_{T}-p=T\left(\frac{\partial p}{\partial T}\right)_{V}-p
$$
On the other hand $\quad p=\frac{R T}{V-b}-\frac{a}{V^{2}}$
or,
$$
T\left(\frac{\partial p}{\partial T}\right)_{V}=\frac{R T}{V-b} \text { and }\left(\frac{\partial U}{\partial V}\right)_{t}=\frac{a}{V^{2}}
$$
So, $\quad \Delta U=a\left(\frac{1}{V_{1}}-\frac{1}{V_{2}}\right)$
(b) From the first law
$$
Q=A+\Delta U=R T \ln \frac{V_{2}-b}{V_{1}-b}
$$