When an elastic spring is stretched or compressed,does its potential energy increase or decrease?

The spring is initially in an unstretched condition. It is first stretched by a length $x$ and the work done is $W_1$. Then again it is stretched by a further length $x$ such that the total extension becomes $2x$. The work done in this second step is $W_2$. The value of $W_2$ is given by

$A$ spring has a spring constant $200 \,N/m$. If it is stretched by $1 \,cm$, then the potential energy stored in it is: (in $\,J$)

$A$ bungee jumper is jumping with the help of an elastic ideal rope (force constant $K$). The jumper steps off the bridge and falls from rest towards the river below. He does not hit the water. The mass of the jumper is $m$,and the natural length of the rope is $l$. Gravity is $g$. Assuming everything is ideal,choose the incorrect option.

$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by a length $L$. The maximum momentum of the block during the collision process is

$A$ spring of original length $L$ and spring constant $K$ is stretched by a small length $x$. It is further stretched by another small length $y$. Find the work done in the second stretching.