$1 .\left(0, \frac{\pi}{6}\right)$
Analysis: Since $f^{\prime}(x)=-\frac{2}{(x-2)^{3}}-2 \geq 0 \Leftrightarrow x \geq 1$, hence $f(x)_{\min }=f(1)$.
We only need $f(1)=\cos 2 \theta-3 \sin \theta+1>0$, which gives $\sin \theta<\frac{1}{2}$, i.e., $\theta \in\left(0, \frac{\pi}{6}\right)$.
1. Given the function $f(x)=\frac{1}{(x-2)^{2}}-2 x+\cos 2 \theta-3 \sin \theta+2$ is always positive for $x \in(-\infty, 2)$. Then the range of the parameter $\theta$ in the interval $(0, \pi)$ is $\qquad$