13. Given that when $x \in[0,1]$, the inequality $x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta>0$ always holds. Try to find the range of $\theta$.
$$
\begin{array}{l}
\text { Let } f(x)=x^{2} \cos \theta-x(1-x)+(1-x)^{2} \sin \theta \\
=(\cos \theta+\sin \theta+1) x^{2}-(2 \sin \theta+1) x+\sin \theta, \\
\text { By the problem, we have }\left\{\begin{array}{l}
f(0)=\sin \theta>0, \\
f(1)=\cos \theta>0
\end{array} \Rightarrow \theta \in\left(2 k \pi, 2 k \pi+\frac{\pi}{2}\right), k \in \mathbf{Z}\right. \text {. }
\end{array}
$$

At this point, the axis of symmetry of $f(x)$ is $x=\frac{2 \sin \theta+1}{2(\cos \theta+\sin \theta+1)} \in(0,1)$, so $f(x)>0$ always holds
$$
\begin{array}{l}
\Rightarrow \Delta=(2 \sin \theta+1)^{2}-4 \sin \theta(\cos \theta+\sin \theta+1)=1-2 \sin 2 \theta\frac{1}{2} \Rightarrow 2 \theta \in\left(2 k \pi+\frac{\pi}{6}, 2 k \pi+\frac{5 \pi}{6}\right) \Rightarrow \theta \in\left(k \pi+\frac{\pi}{12}, k \pi+\frac{5 \pi}{12}\right) .
\end{array}
$$

In summary, the range of $\theta$ is $\left(2 k \pi+\frac{\pi}{12}, 2 k \pi+\frac{5 \pi}{12}\right), k \in \mathbf{Z}$.
$$