The potential energy of a long spring when stretched by 2 cm  is U. If the spring is stretched by 8 cm , potential energy stored in it will be:

Elastic potential energy stored in the spring

$U=\frac{1}{2}k{x}^2$

$=\frac{1}{2}Fx=\frac{F^2}{2k}$

$Hence\frac{U_1}{U_2}=\frac{k_2}{k_1}\left. [Fissameforbothsprings\right]$

$=\frac{500}{1200}=\frac{5}{12}$

ME = PE + KE = 8J
At x? , PE=4J, so KE = 8-4=4J. (A is correct)
At x>x? , PE=6J, so KE=2J, which is constant. (B is correct)
At x8J. This is not possible. Particle cannot reach here. So it is not a correct statement. (C is incorrect)
At x? , PE=0J, so KE=8J, which is maximum. (D is correct)