I suppose that the spring follows Hooke's Law:
where `F` is the force (it is given), `k` is the spring force constant (given also) and `x` is the length of the elongation or compression. Using this formula we can find the length of an elongation,
Also it is known that for such a springs the elastic potential energy is equal to
Now perform the calculations.
For the first case, the elongation `x` is `F/k=375/525 approx 0.71(m).`
The elastic potential energy is `(375^2)/(2*525) approx 134 (J).`
For the second case, the elongation x is `F/k=350/525 approx 0.67(m).`
The elastic potential energy is `(350^2)/(2*525) approx 117 (J).`
There is no surprise that for the less force the elongation occurred to be less also, and the same is observed for the elastic potential energy.