复数 $z$ 的极坐标形式由 $z = r\text{e}^{\text{i}\theta}$ 给出，其中 $r > 0$ 且 $0 < \theta \le \pi$，复数 $w = \left(\frac{1}{3} - \frac{\sqrt{3}}{3}\text{i}\right)z$。

1. Find the modulus and argument of the complex number multiplier $\frac{1}{3} - \frac{\sqrt{3}}{3}\text{i}$.

Let $c = \frac{1}{3} - \frac{\sqrt{3}}{3}\text{i}$.
$|c| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(-\frac{\sqrt{3}}{3}\right)^2} = \sqrt{\frac{1}{9} + \frac{3}{9}} = \frac{2}{3}$
Since the real part is positive and the imaginary part is negative, the complex number lies in the fourth quadrant.
$\arg(c) = \tan^{-1}\left(\frac{-\sqrt{3}/3}{1/3}\right) = \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$
Thus, $c = \frac{2}{3}\text{e}^{-\text{i}\frac{\pi}{3}}$.

2. Multiply the complex numbers to express $w$ in exact polar form.

$w = c \times z = \left(\frac{2}{3}\text{e}^{-\text{i}\frac{\pi}{3}}\right)\left(r\text{e}^{\text{i}\theta}\right)$
$w = \frac{2}{3}r\text{e}^{\text{i}\left(\theta - \frac{\pi}{3}\right)}$
Since $0 < \theta \le \pi$, the argument $\theta - \frac{\pi}{3}$ lies within the principal range $(-\pi, \pi]$, so no adjustment is required.

1. Express $z^5$ and $w^*$ in exponential form.

From part (i), $w = \frac{2}{3}r\text{e}^{\text{i}\left(\theta - \frac{\pi}{3}\right)}$.
The complex conjugate $w^*$ is found by negating the argument:
$w^* = \frac{2}{3}r\text{e}^{-\text{i}\left(\theta - \frac{\pi}{3}\right)} = \frac{2}{3}r\text{e}^{\text{i}\left(\frac{\pi}{3} - \theta\right)}$
Using standard index laws (De Moivre's Theorem for exponential forms),
$z^5 = (r\text{e}^{\text{i}\theta})^5 = r^5\text{e}^{\text{i}5\theta}$

2. Find the quotient $\frac{z^5}{w^*}$ in exponential form.

$\frac{z^5}{w^*} = \frac{r^5\text{e}^{\text{i}5\theta}}{\frac{2}{3}r\text{e}^{\text{i}\left(\frac{\pi}{3} - \theta\right)}}$
$= \frac{3}{2}r^4 \text{e}^{\text{i}\left(5\theta - \left(\frac{\pi}{3} - \theta\right)\right)}$
$= \frac{3}{2}r^4 \text{e}^{\text{i}\left(6\theta - \frac{\pi}{3}\right)}$

3. Form an equation using the condition that the quotient is real and positive.

For a complex number to be real and positive, its imaginary part is 0 and its real part is positive, which means its argument must be an integer multiple of $2\pi$.
$6\theta - \frac{\pi}{3} = 2k\pi, \quad \text{for } k \in \mathbb{Z}$
$6\theta = 2k\pi + \frac{\pi}{3} = \frac{(6k + 1)\pi}{3}$
$\theta = \frac{(6k + 1)\pi}{18}$

4. Solve for the possible values of $\theta$ within the specified domain.

We are given the domain $0 < \theta \le \pi$.
Substituting integer values for $k$:
For $k = 0$, $\theta = \frac{\pi}{18}$
For $k = 1$, $\theta = \frac{7\pi}{18}$
For $k = 2$, $\theta = \frac{13\pi}{18}$
For $k = 3$, $\theta = \frac{19\pi}{18}$ (Reject, as $\theta > \pi$)
For $k < 0$, $\theta < 0$ (Reject, as $\theta \le 0$)

The possible exact values are $\frac{\pi}{18}$, $\frac{7\pi}{18}$, and $\frac{13\pi}{18}$.