1. Given the complex number $z_{1}=2+\mathrm{i}, 2 z_{2}=$ $\frac{z_{1}+\mathrm{i}}{(2 \mathrm{i}+1)-z_{1}}$. If the three interior angles $\angle A$, $\angle B$, and $\angle C$ of $\triangle A B C$ form an arithmetic sequence, and $u=\cos A+$ $2 \mathrm{i} \cos ^{2} \frac{C}{2}$, then the range of $\left|u+z_{2}\right|$ is $\qquad$.
II. $1 \cdot\left[\frac{\sqrt{2}}{2}, \frac{\sqrt{5}}{2}\right)$.
From the problem, we know that $z_{2}=-\mathrm{i}, \angle B=60^{\circ}, \angle A+\angle C=120^{\circ}$, then
$$
u+z_{2}=\cos A+\left(2 \cos ^{2} \frac{C}{2}-1\right) \mathrm{i}=\cos A+\mathrm{i} \cos C .
$$

Therefore, $\left|u+z_{2}\right|^{2}=\cos ^{2} A+\cos ^{2} C$
$$
\begin{array}{l}
=1+\frac{1}{2}(\cos 2 A+\cos 2 C) \\
=1+\cos (A+C) \cdot \cos (A-C) \\
=1-\frac{1}{2} \cos (A-C) .
\end{array}
$$

Since $\angle A-\angle C=120^{\circ}-2 \angle C, 0^{\circ}<\angle C<120^{\circ}$,
then $-120^{\circ}<\angle A-\angle C<120^{\circ}$.
Thus, $-\frac{1}{2}<\cos (A-C) \leqslant 1$.
Therefore, $\frac{\sqrt{2}}{2} \leqslant\left|u+z_{2}\right|<\frac{\sqrt{5}}{2}$.