Answer
(a) There is complete transfer of energy for $m_1=m_2$.
Work Step by Step
Kinetic energy is given by $K=\frac{p^2}{2m}$.
Initial kinetic energy for this elastic collision, $K_i=\frac{p^2}{2m_1}$
For a $m_1$ hitting a stationary $m_2$ and complete transfer of energy, Final kinetic energy is $K_f=\frac{p^2}{2m_2}$, since momentum is conserved.
From the second condition for elastic collision -
\begin{align*}
K_i&=K_f\\
or,\, \frac{p^2}{2m_1}&=\frac{p^2}{2m_2}\\
or,\, m_1 &=m_2
\end{align*}