$x(t) = A\sin(\omega t + \phi)$

$v(t) = A\omega \cos(\omega t + \phi)$

$x(0) = -\sqrt{3}$, $v(0) = 3$, $\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{9}{1}} = 3$, $\phi = -\dfrac{\pi}{3}$

using the above information gives

$x(t) = 2\sin \left(3 t - \dfrac{\pi}{3} \right)$

$v(t) = 6\cos \left(3 t - \dfrac{\pi}{3}\right)$

passing through equilibrium, $x(t) = 0 \implies 3t - \dfrac{\pi}{3} = \{0, \pi, 2\pi, ... \} \implies t = \bigg\{ \dfrac{\pi}{9}, \dfrac{4\pi}{9}, \dfrac{7\pi}{9}, ... \bigg \}$

the {1st, 3rd, 5th, ...} passes through equilibrium are in the upward direction

(h) set $x(t) = -1$ and solve for time

(i) use solutions from (h) and evaluate if v(t) > 0

(j) use solutions from (h) and evaluate if v(t) < 0