Answer
See the detailed answer below.
Work Step by Step
a) The system [glider+spring] here undergoes a simple harmonic motion.
We know that the initial position of the air-track glider, at $t=0$, is given by
$$x_0=A\cos\phi_0\tag 1$$
and hence the initial velocity is given by
$$v_{0x}=-\omega A\sin\phi_0\tag2$$
Dividing (2) by (1);
$$\dfrac{v_{0x}}{x_0}=\dfrac{-\omega \color{red}{\bf\not}A\sin\phi_0}{ \color{red}{\bf\not}A\cos\phi_0}=-\omega \tan\phi_0$$
Thus,
$$\phi_0=\tan^{-1}\left[\dfrac{v_{0x}}{-\omega x_0}\right]\tag 3$$
where $\omega=2\pi f$ and $f=1/T$, so
$$\omega=\dfrac{2\pi}{T}$$
Plugging into (3);
$$\phi_0=\tan^{-1}\left[\dfrac{Tv_{0x}}{-2\pi x_0}\right] $$
We know that $x_0=-5$ cm since it starts from the left of the equilibrium point and moves right, $v_{0x}=36.3$ cm/s, $T=1.5$ s.
Plugging the known and recalling that this is a radian angle not in degrees, you need to be careful when using your calculator.
$$\phi_0=\tan^{-1}\left[\dfrac{(1.5)(36.3)}{-2\pi (-5)}\right]=1.0475\;\rm rad \approx 60^\circ$$
Noting that our object is initially assumed to be on the third quadrant since it starts from a negative position but moves to the right.
Thus, the phase constant is given by
$$\phi_0=60^\circ+180^\circ={\bf 240^\circ}$$
which is as same as
$$\phi_0=\color{red}{\bf -120^\circ}=\color{red}{\bf \frac{-2\pi}{3}}\; \rm rad$$
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b) We know that
$$\phi=\omega t+\phi_0=2\pi f t+\phi_0=\dfrac{2\pi t}{T}+\phi_0$$
$\rightarrow$ At $t=0$ s, which we found above.
$$\phi=\dfrac{2\pi (0)}{T}+\phi_0=\phi_0=\color{red}{\bf \frac{-2\pi}{3}}\; \rm rad$$
$\rightarrow$ At $t=0.5$ s,
$$\phi=\dfrac{2\pi (0.5)}{1.5}+\frac{-2\pi}{3}=\color{red}{\bf 0}\; \rm rad$$
$\rightarrow$ At $t=1$ s,
$$\phi=\dfrac{2\pi (1)}{1.5}+\frac{-2\pi}{3}=\color{red}{\bf \frac{ 2\pi}{3}}\; \rm rad$$
$\rightarrow$ At $t=1.5$ s, which is after one full cycle
$$\phi=\dfrac{2\pi (1.5)}{1.5}+\frac{-2\pi}{3}=\color{red}{\bf \frac{ 4\pi}{3}}\; \rm rad$$
