A block is suspended by an ideal spring of force constant k. If the block is pulled down by applying a constant force F and if maximum displacement of the block from its initial position of rest is $\mathrm{\delta },$ then 

Let the mass of the block hanging from the spring be m. Then initial elongation of the spring will be equal to mg/k. When the force F is applied to pull the block down, the work done by F and further loss of gravitational potential energy of the block is used to increase strain energy of this spring.

         $(\mathrm{F\delta }+\mathrm{mg\delta })=\left[\frac{1}{2}\mathrm{k}{\left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{\delta }\right)}^2-\frac{{\mathrm{m}}^2{\mathrm{g}}^2}{2\mathrm{k}}\right]$                          …(i)

From Eq.(i) $\mathrm{\delta }=2\mathrm{F}/\mathrm{k}$

Hence, option (2) is correct.

Since a constant force is applied on the block to pull it down, during this process, work done by force, $\mathrm{W}=\mathrm{F\delta }$. Hence, option (3) is also correct.

Increase in energy of the spring is equal to

         $\left[\frac{1}{2}\mathrm{k}{\left(\frac{\mathrm{mg}}{\mathrm{k}}+\mathrm{\delta }\right)}^2-\frac{{\mathrm{m}}^2{\mathrm{g}}^2}{2\mathrm{k}}\right]$

From this equation, increase in energy is not equal to $\frac{1}{2}{\mathrm{k\delta }}^2$.

Hence, option (4) is wrong.