A spring stretches by 2 mm when it is loaded with a mass of 200 g....

At equilibrium, the restoring force of the spring balances the weight of the mass:

$mg = kx_0$

$k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000$ N/m

The frequency of oscillation is given by:

$f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi} \sqrt{\dfrac{1000}{0.2}} = \dfrac{1}{2\pi} \sqrt{5000} = \dfrac{10\sqrt{50}}{2\pi} = \dfrac{5\sqrt{50}}{\pi}$ Hz

The maximum energy in the spring corresponds to its maximum elastic potential energy, which occurs at the maximum extension.

Maximum extension x_\max = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3} m

Maximum energy in the spring U_\max = \dfrac{1}{2} k x_\max^2

U_\max = \dfrac{1}{2} \times 1000 \times (4 \times 10^{-3})^2

U_\max = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3} J

Answer: $\dfrac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$