A platform of mass 0.30 kg sits on top of a vertical spring of 200 N/m…

(a) $\omega = \sqrt{\dfrac{200\ \text{N/m}}{0.50\ \text{kg}}} = 20\ \text{rad/s}$.

(b) At the top the acceleration is $\omega^2A$, downward. Gravity alone can give the block $g$ downward. If the motion needs more than $g$, the platform would have to pull the block down, which it cannot do. So $\omega^2A \le g$, and $A_{max} = \dfrac{9.8\ \text{m/s}^2}{(20\ \text{rad/s})^2} = 0.0245\ \text{m} = 2.45\ \text{cm}$.

(c) $E = \tfrac12 \times 200\ \text{N/m} \times (0.0245\ \text{m})^2 = 0.060\ \text{J}$ and $v_{max} = 20\ \text{rad/s} \times 0.0245\ \text{m} = 0.49\ \text{m/s}$.

Why at the top? At the bottom the acceleration points up, so the platform pushes harder on the block. Contact is lost only near the top.