$hf-{{W}_{\circ }}={{K}_{\max }}\Rightarrow \frac{hc}{{{\lambda }_{1}}}-4=2\Rightarrow \frac{hc}{{{\lambda }_{1}}}=6\Rightarrow \frac{hc}{{{\lambda }_{2}}}=\frac{hc}{\frac{3}{4}{{\lambda }_{1}}}=\frac{4}{3}\times 6=8eV$ $K{{'}_{\max }}=\frac{hc}{{{\lambda }_{2}}}-{{W}_{\circ }}=8-4=4eV$