1. Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the range of the angle between its two asymptotes is $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, then the range of the eccentricity of the hyperbola is $\qquad$
1. $\left[\frac{2 \sqrt{3}}{3}, 2\right]$ Detailed explanation: Let the angle between the two asymptotes be $2 \theta$. Since the eccentricity $e^{2}=\left(\frac{c}{a}\right)^{2}=\frac{a^{2}+b^{2}}{a^{2}}=1+\left(\frac{b}{a}\right)^{2}$, from the problem, when $\theta \in\left(0, \frac{\pi}{4}\right], k=\frac{b}{a}=\tan \theta, 2 \theta \in\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, then it is easy to know $\theta \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$, so $k=\frac{b}{a}=\tan \theta \in$ $\left[\frac{\sqrt{3}}{3}, \sqrt{3}\right], e^{2} \in\left[\frac{4}{3}, 4\right]$, hence $e \in\left[\frac{2 \sqrt{3}}{3}, 2\right]$; when $\theta \in\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, it is easy to know $e \in\left[\frac{2 \sqrt{3}}{3}, 2\right]$.