Q.A spring which is initially in an unstretched condition is first stretched by a length $x$, and again by a further length $x$. The work done in the first case $W_1$ is one third of the work done in the second case $W_2$. True or false?

A stretched or compressed spring stores elastic potential energy. For a spring of force constant $k$ obeying Hooke's law, the internal restoring force is $\vec F_s=-k\vec x$ (the negative sign showing it always opposes the displacement $x$ from the natural, equilibrium length), so the external force needed to produce and hold a stretch $x$ is the equal-and-opposite, position-dependent force $F_a=kx$.

Because this applied force is variable (proportional to $x$, unlike the constant gravitational force), the stored energy must be found by integrating it from the equilibrium position (where $U=0$) out to the final extension $x$:

If the spring starts at some nonzero extension $x_i$ and is stretched further to $x_f$, the more general result is $U=\tfrac12k(x_f^2-x_i^2)$. Two properties of this stored energy are worth remembering: it depends only on the spring constant $k$ and the displacement $x$ -- never on the mass attached to the spring -- and it equals the triangular area under the straight-line $F$-versus-$x$ graph of the spring force. In a frictionless spring-mass system, this elastic potential energy trades back and forth with the attached mass's kinetic energy, being maximum (and purely potential) at the extreme stretched/compressed positions and zero (with kinetic energy at its maximum) as the mass crosses the equilibrium position -- their sum, the total mechanical energy, staying constant throughout, exactly as required by the law of conservation of energy.