Answer
(a) The magnitude of acceleration when the extension of the spring is maximum is $9.80~m/s^2$.
(b) The maximum extension of the spring is $0.784~m$
Work Step by Step
(a) Initially, the acceleration is $9.80~m/s^2$ downward. Because the body is in SHM, the body will have the same magnitude of acceleration when the extension of the spring is maximum. The magnitude of acceleration when the extension of the spring is maximum is $9.80~m/s^2$.
(b) We can find the maximum extension of the spring $d$:
$\sum F = m~a_y$
$kd-mg = m~a_y$
$kd = m~a_y+mg$
$d = \frac{m~(a_y+g)}{k}$
$d = \frac{(1.0~kg)~(9.80~m/s^2+9.80~m/s^2)}{25~N/m}$
$d = 0.784~m$
The maximum extension of the spring is $0.784~m$
