15-59.

The solution for a damped spring can be written in the form

$$x(t)=A(t)\cos(\omega t+\varphi)$$

where we have

$$A(t)=Ae^{-bt/2m}$$

When written in this form it is obvious that the non-damped spring is written the same way with $A(t)=A$ is time-independent. In any case, part a) refers to the amplitude only - the $A(t)$. Thus, since

$$A_0=A(0)=A$$

we have

$$\begin{eqnarray*} A/3 &=& A(t)=Ae^{-bt/2m}\\ 3^{-1} &=& e^{-bt/2m}\\ -\ln 3 &=& -\frac{bt}{2m}\\ t &=& 2\ln 3 \frac{m}{b}\\ &=& 14.33 \rm {s}. \end{eqnarray*}$$

There are a few ways to consider the next part. If there are $n$ oscillations then the angle has changed by an mount $2\pi n$, that is

$$\Delta\theta=2\pi n = \omega_f-\omega_i=(\omega t +\varphi) - (\varphi) = \omega t.$$

Which yields

$$n=\frac{\omega t}{2\pi}=ft=\frac{t}{T},$$

expressed in various ways (note that we don't need $\varphi$). Thus all we need is $\omega$ for a damped system, which is given by

$$\omega=\sqrt{\frac{k}{m}-\left(\frac{b}{2m}\right)^2}.$$