To determine the amplitude and phase of the wave formed by the superposition of the two harmonic traveling waves, $ y_1(x, t) = 4 \sin (kx - \omega t) $ and $ y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3}) $, we use the following steps:

First, calculate the resultant amplitude $ A $ using the formula for the resultant of two waves with different phases:

$ A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi)} $

where:

$ A_1 = 4 $ (amplitude of the first wave)

$ A_2 = 2 $ (amplitude of the second wave)

$ \phi = 120^\circ $ (phase difference, given by $\frac{2\pi}{3}$).

Substituting the values, we get:

$ A = \sqrt{4^2 + 2^2 + 2 \times 4 \times 2 \times \cos(120^\circ)} $

$ \cos(120^\circ) = -\frac{1}{2} $

Therefore:

$ A = \sqrt{16 + 4 + 16 \times (-\frac{1}{2})} $

$ A = \sqrt{16 + 4 - 8} $

$ A = \sqrt{12} = 2\sqrt{3} $

Next, compute the phase $ \phi $ of the resultant wave:

$ \tan \phi = \frac{A_2 \sin(\phi)}{A_1 + A_2 \cos(\phi)} $

$ \tan \phi = \frac{2 \sin(120^\circ)}{4 + 2 \cos(120^\circ)} $

$ \sin(120^\circ) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \cos(120^\circ) = -\frac{1}{2} $

Substituting these values, we get:

$ \tan \phi = \frac{2 \times \frac{\sqrt{3}}{2}}{4 + 2 \times -\frac{1}{2}} $

$ \tan \phi = \frac{\sqrt{3}}{3} $

This simplifies to:

$ \phi = \frac{\pi}{6} $

Thus, the amplitude and phase of the resultant wave are $ 2\sqrt{3} $ and $ \frac{\pi}{6} $ respectively.