Answer
(a) $k = 3540~N/m$
(b) $T = 0.851~s$
(c) $v_{max} = 0.369~m/s$
Work Step by Step
(a) We can find the force constant of the spring.
$kx = mg$
$k = \frac{mg}{x}$
$k = \frac{(65.0~kg)(9.80~m/s^2)}{0.180~m}$
$k = 3540~N/m$
(b) We can find the period of oscillation.
$T = 2\pi~\sqrt{\frac{m}{k}}$
$T = 2\pi~\sqrt{\frac{65.0~kg}{3540~N/m}}$
$T = 0.851~s$
(c) We can find the maximum speed.
$v_{max} = A~\omega$
$v_{max} = A~\sqrt{\frac{k}{m}}$
$v_{max} = (0.0500~m)~\sqrt{\frac{3540~N/m}{65.0~kg}}$
$v_{max} = 0.369~m/s$
