Explanation

(c) The particles are moving with velocities 0.8 $c$ and $- 0.4\ c$ in the laboratory frame, say $S'$ frame. Let $S$ be the reference frame in which the particle with velocity $- 0.4\ c$ is at rest. Then, the velocity of $S'$ (laboratory) relative to $S$ is $v = 0.4\ c.$ Therefore, the particle which is $S'$ has velocity $u^{'} = + 0.8\ c$ has a velocity in $S$ given by
$u = \frac{u^{'} + v}{1 + \frac{u^{'}v}{c^{2}}} = \frac{0.8\ c + 0.4\ c}{1 + \frac{(0.8c)(0.4c)}{c^{2}}} = \frac{1.2c}{1 + 0.32} = 0.9c$