A block oscillates on a spring with an amplitude of 10 cm…

(a) $K = 3U$ means $U = \tfrac14 E$. So $x^2 = \tfrac14 A^2$, and $x = \dfrac{10\ \text{cm}}{2} = 5.0\ \text{cm}$.

(b) $U = \tfrac12 E$ gives $x = \dfrac{10\ \text{cm}}{\sqrt2} = 7.07\ \text{cm}$.

(c) There $K = \tfrac12 E$. Since $K$ goes as $v^2$, $v = \dfrac{v_{max}}{\sqrt2} = 0.707\,v_{max}$.