The conservation laws of (classical kinetic) energy and (linear) momentum determine the outcome of the collision. The final speed of the $$\alpha$$ particle is
$$ v_{\alpha f} = \dfrac {m_\alpha - m_{Au}} {m_\alpha + m_{Au}} v_{\alpha i}. $$
and that of the recoiling gold nucleus is
$$ v_{Au, f} = \dfrac {2m_\alpha}{m_\alpha + m_{Au}} v_{\alpha i} $$
(a) Therefore, the kinetic energy of the recoiling nucleus is
$$ K_{Au, f} = \dfrac 12 m_{Au} v^2_{Au,f} = \dfrac 12 m_{Au} \left(\dfrac {2m_\alpha}{m_\alpha + m_{Au}} \right)^2 v^2_{\alpha i} = K_{\alpha i} \dfrac{4m_{Au} m_{\alpha}} {(m_{\alpha} + m_{Au})^2} $$
$$ = (5.00 MeV)\dfrac{4(197 u)(4.00 u)} {(4.00u + 197 u)^2}$$
$$ = 0.390 MeV $$
(b) The final kinetic energy of the alpha particle is
$$K_{\alpha f} = \dfrac 12 m_{\alpha} v^2_{\alpha f} = \dfrac 12 m_{\alpha} \left(\dfrac {m_{\alpha} - m_{Au}} {m_{\alpha} + m_{Au}} \right)^2 v^2_{\alpha i} = K_{\alpha i} \left(\dfrac {m_{\alpha} - m_{Au}} {m_{\alpha} + m_{Au}} \right) ^2 $$
$$= (5.00 MeV) \left (\dfrac {4.00u - 197u} {4.00u + 197u} \right)^2 $$
$$= 4.61 MeV. $$
We note that $$K_{af} + K_{Au, f} = K_{\alpha i} $$ is indeed satisfied.