Part (a):
$$\begin{equation*}
u(x,t) = U(x - vt, t)\\
Let \xi = x - vt, \eta = t \\
u_x = U_x = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial x} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial x} = U_x\\
u_{xx} = U_{\xi\xi}\\
u_t = U_t = \frac{\partial U}{\partial \xi}\frac{\partial \xi}{\partial t} + \frac{\partial U}{\partial \eta}\frac{\partial \eta}{\partial t} = -vU_{\xi} + U_{\eta}\\
-vU_{\xi} + U_{\eta} + vU_{\xi} = kU_{\xi\xi} \\
U_{\eta} = kU_{\xi\xi}\\
U(\xi,\eta) = \int_{-\infty}^{\infty}G(\xi,y,\eta)g(y)dy\\
u(x,t) = \int_{-\infty}^{\infty}G(x - vt,y,t)g(y)dy \\
= \int_{-\infty}^{\infty} \frac{1}{2 \sqrt{k \pi t}} e^{\frac{-(x-vt-y)^2}{4kt}}g(y)dy
\end{equation*}$$
Part (b):
Let $$
\begin{equation*}
u(x,t) = e^{\alpha t + \beta x}V(x,t)\\
u_t = \alpha e^{\alpha t + \beta x}V + e^{\alpha t + \beta x}V_t\\
u_x = \beta e^{\alpha t + \beta x} + e^{\alpha t + \beta x}V_t\\
u_{xx} = \beta^2 e^{\alpha t + \beta x}V + 2\beta e^{\alpha t + \beta x}V_x + e^{\alpha t + \beta x}V_{xx}\\
\end{equation*}$$
Now plug these into the given heat equation.
$$
\begin{equation*}
\alpha V + V_t + v\beta V + vV_x = k\beta^2 V + 2k\beta V_x + kV_{xx} \end{equation*}$$
set $ \alpha + v\beta = k\beta^2$ and $ v = 2k\beta$ then,
$$\begin{equation*}
\beta = \frac{v}{2k}, \alpha = \frac{-v^2}{4k} \\
u(x,t) = e^{\alpha t + \beta x} \int_{0}^{\infty} G_D(x,y,t)g(y)dy\end{equation*} \\$$
This works because if $u(0,t) = 0,$ then $ V(0,t) = 0.$
