Let b denotes boy and m denotes man.

Case 1:

A running man has the same kinetic energy as that of a boy of half his mass.

$(\frac{1}{2}{m_{m1}}v_{m1}^2 = \frac{1}{2}{m_{b1}}v_{b1}^2 \Rightarrow {m_{m1}}v_{m1}^2 = {m_{b1}}v_{b1}^2)$

$(m \times v_{m1}^2 = \frac{m}{2} \times v_{b1}^2 \Rightarrow {v_{b1}} = \sqrt 2 {v_{m1}})$

Case 2:
The man speeds up by 2 m/s and the boy changes his speed up by x m/s

$(\frac{1}{2}{m_{m2}}v_{m2}^2 = \frac{1}{2}{m_{b2}}v_{b2}^2)$

$(m \times v_{m2}^2 = \frac{m}{2} \times v_{b2}^2 \Rightarrow {v_{b2}} = \sqrt 2 {v_{m2}})$

$(\left( {{v_{b1}} + x} \right) = \sqrt 2 \left( {{v_{m1}} + 2} \right) = \sqrt 2 {v_{m1}} + 2\sqrt 2)$

$(\sqrt 2 {v_{m1}} + x = \sqrt 2 {v_{m1}} + 2\sqrt 2 \Rightarrow x = 2\sqrt 2)$