Answer
See the detailed answer below.
Work Step by Step
a) We know that the frequency of an object that undergoes a simple harmonic motion is given by
$$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$$
where $k$ is the spring constant and $m$ is the object's mass.
So the spring constant is found by squaring both sides and then solving for $k$;
$$4\pi^2 f^2=\dfrac{k}{m}$$
$$k=4\pi^2 f^2 m$$
Plugging the known;
$$k=4\pi^2 (4.0)^2 (15\times 10^{-3})$$
$$k=\color{red}{\bf 9.475}\;\rm N/m$$
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b) In damping oscillations, we know that the amplitude is given by
$$x_{(t)}=A_0e^{-t/2\tau}$$
where $\tau= m/b$, so
$$x_{(t)}=A_0e^{-bt/2m}$$
where $b$ is the damping constant we need to find.
So,
$$\dfrac{x_{(t)}}{A_0}=e^{-bt/2m}$$
$$\ln\left[ \dfrac{x_{(t)}}{A_0} \right]=\dfrac{-bt}{2m}$$
Hence,
$$b=-\left(\dfrac{2 m}{t}\right)\ln\left[ \dfrac{x_{(t)}}{A_0} \right] \tag 1$$
The author here did not tell us the initial amplitude so we can't find the magnitude of $b$. But yet we can plug the known;
$$ b=-\left(\dfrac{ 2(15\times 10^{-3})}{4}\right)\ln\left[ \dfrac{(0.5\times 10^{-2})}{A_0} \right] $$
Let's assume that $A_0=4$ cm, so
$$ b=-\left(\dfrac{ 30\times 10^{-3}}{4}\right)\ln\left[ \dfrac{(0.5\times 10^{-2})}{(4\times 10^{-2})} \right] $$
$$b=\bf 15.6\times 10^{-3}\;\rm kg/s$$
