Answer
$ a.\quad$shown below
$ b.\quad$shown below
Work Step by Step
$(y$ is proportional to $x $ if there is a constant $k$ such that $y=kx)$
$ a.\quad$
$s(t)=a\cos kt+b\sin ktx\qquad $... differrentiate (chain rule, twice)
$\displaystyle \frac{ds}{dt}=-a\sin kt\cdot k+b\cos kt\cdot k$
$\displaystyle \frac{ds}{dt}=-ak\sin kt+bk\cos kt\qquad $... differrentiate (chain rule, twice)
$\displaystyle \frac{d^{2}s}{dt^{2}}=-ak^{2}\cos kt-bk^{2}\sin kt$
$=-k^{2} (a \cos kt+b\sin kt)$
$=(-k^{2})\cdot s$
So, $\displaystyle \frac{d^{2}s}{dt^{2}}$ is proportional to $s$.
The negative sign indicates that the acceleration is in the opposite direction of displacement (towards the origin).
$ b.\quad$
$s(t)=a\cosh kt-\vdash b\sinh kt \qquad $... differrentiate
$\displaystyle \frac{ds}{dt}=ak\sinh kt+bk\cosh kt \qquad $... once more,
$\displaystyle \frac{d^{2}s}{dt^{2}}=ak^{2}\cosh kt+bk^{2}\sinh kt=k^{2}s$
Again, acceleration is proportional to displacement.
This time, acceleration has the same sign (direction) as displacement (away from the origin).
