#### Answer

(a) $\phi_0 = \frac{5\pi}{3}$
(b) The phase at point 1 is $\frac{4\pi}{3}$
The phase at point 2 is $0$
The phase at point 3 is $\frac{2\pi}{3}$

#### Work Step by Step

(a) The general equation for the motion of the particle is:
$x(t) = A~cos(\omega~t+\phi_0)$
We can find $\phi_0$ when $x(0) = \frac{A}{2}$
$cos(\phi_0) = \frac{1}{2}$
$\phi_0 = arccos(\frac{1}{2})$
$\phi_0 = \frac{\pi}{3}, \frac{5\pi}{3}$
In the graph, we can see that a standard cos-curve has been shifted to the left by an angle of $\frac{5\pi}{3}$. Therefore, the phase constant is $\phi_0 = \frac{5\pi}{3}$.
(b) We can find the phase at point 1:
$phase = \pi+\frac{\pi}{3} = \frac{4\pi}{3}$
We can find the phase at point 2:
$phase = 0$
We can find the phase at point 3:
$phase = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$