Answer
$\Delta \phi= \dfrac{2 \pi}{3}$
Work Step by Step
The two displacements equal half of the amplitude so could get the next
$$\cos \left(\omega t+\phi_{1}\right)=\cos \left(\omega t+\phi_{2}\right)=\frac{1}{2}$$
Now take $\cos^{-1}$ to get the phase by $\pi/3$. We are given that the two particles are in opposite moving, so both phases have opposite sign, therefore, the phase difference between both particles will be
$$\Delta \phi=\frac{\pi}{3}-\left(\frac{-\pi}{3}\right)=\frac{2 \pi}{3}$$
