Start off with differentiating
![](https://latex.codecogs.com/png.latex?\bg_white x = b + 2b cos (nt))
with respect to time until you get to acceleration. Then put acceleration in the form
![](https://latex.codecogs.com/png.latex?\bg_white -n^2(x-c))
to assist with substitution.
From the question, when:
![](https://latex.codecogs.com/png.latex?\bg_white x = 3b)
,
and
![](https://latex.codecogs.com/png.latex?\bg_white x = 2b)
,
Substitute the conditions into the relevant equations to get two simultaneous equations in terms of
![](https://latex.codecogs.com/png.latex?\bg_white b)
and
![](https://latex.codecogs.com/png.latex?\bg_white n)
. Solve for
![](https://latex.codecogs.com/png.latex?\bg_white b)
and
![](https://latex.codecogs.com/png.latex?\bg_white n)
and divide
![](https://latex.codecogs.com/png.latex?\bg_white 2\pi)
by
![](https://latex.codecogs.com/png.latex?\bg_white n)
to get the period of motion.
For 2, know that maximum speed occurs at the centre of motion. Rearrange the displacement equation and use some basic trig to get the final result.