Answer
(a) $\sqrt{v_{\circ}^2+\frac{KQ^2}{ma}(-4+\sqrt{2})}$
(b) less than
Work Step by Step
(a) According to the law of conservation of energy
$K_i+U_i=K_f+U_f$
$\implies \frac{1}{2}mv_{\circ}^2+\frac{KQ^2}{a}(-2+\frac{1}{\sqrt{2}})=\frac{1}{2}mv^2+0$
This simplifies to:
$v=\sqrt{v_{\circ}^2+\frac{KQ^2}{ma}(-4+\sqrt{2})}$
(b) We know that the other negative charge that is $q_3$ is acted upon by a stronger attractive force. Therefore, in order to move away from the $q_2$ and $q_4$ charges, it dissipates more energy and consequently it slows down. We conclude that the final speed will be less than that of the first $-Q$ charge -- that is, $q_1$.
