A glider of 0.50 kg on a spring of 50 N/m oscillates with an amplitude of 12 cm…

(a) $E = \tfrac12 \times 50\ \text{N/m} \times (0.12\ \text{m})^2 = 0.36\ \text{J}$.

(b) At the middle, $\tfrac12 mv_{max}^2 = E$. We want $v_{max}$, so $v_{max} = \sqrt{\dfrac{2 \times 0.36\ \text{J}}{0.50\ \text{kg}}} = 1.2\ \text{m/s}$.

(c) $U = \tfrac12 \times 50\ \text{N/m} \times (0.060\ \text{m})^2 = 0.09\ \text{J}$, so $K = 0.36\ \text{J} - 0.09\ \text{J} = 0.27\ \text{J}$.