1. Let the complex number $z=\cos \theta+\mathrm{i} \sin \theta(0<\theta<\pi), \omega=\frac{1-(\bar{z})^{4}}{1+z^{4}}$, given that $|\omega|=\frac{\sqrt{3}}{3}$, $\arg \omega<\frac{\pi}{2}$, find $\theta$.
1. Given $|z|=1, \bar{z}=1, \omega=\frac{1-(\bar{z})^{4}}{1+z^{4}}$, we have
$$
\begin{aligned}
\omega \bar{\omega} & =\frac{1-(\bar{z})^{4}}{1+z^{4}} \cdot \frac{1-z^{4}}{1+(\bar{z})^{4}}=\frac{1-(\bar{z})^{4}-z^{4}+(z \bar{z})^{4}}{1+(\bar{z})^{4}+z^{4}+(z \bar{z})^{4}}=\frac{2-\left(\bar{z}^{4}+z^{4}\right)}{2+\left(\bar{z}^{4}+z^{4}\right)} \\
& =\frac{2-2 \cos 4 \theta}{2+2 \cos 4 \theta}=\frac{1-\cos 4 \theta}{1+\cos 4 \theta}=\tan ^{2} 2 \theta .
\end{aligned}
$$

Thus, $|\omega|^{2}=\tan ^{2} 2 \theta$, so $\tan 2 \theta= \pm \frac{\sqrt{3}}{3}$. Since $0<\theta<\pi$, we find that $\theta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{7 \pi}{12}$ or $\frac{11 \pi}{12}$.
Given $0 \leqslant \arg \omega \leqslant \frac{\pi}{2}$, upon verification, we get $\theta=\frac{\pi}{12}$, or $\frac{7 \pi}{12}$.