3. $\left[1, \frac{\sqrt{5}+1}{2}\right)$.

Let $q=\min \left\{\frac{\sin \beta}{\sin \alpha}, \frac{\sin \gamma}{\sin \beta}\right\}$.
Then $q \leqslant \frac{\sin \beta}{\sin \alpha}, q \leqslant \frac{\sin \gamma}{\sin \beta}$
$$
\begin{aligned}
\Rightarrow & \frac{1}{q} \geqslant \frac{\sin \alpha}{\sin \beta}, \frac{1}{q} \geqslant \frac{\sin \beta}{\sin \gamma} \\
\Rightarrow & \frac{1}{q^{2}}+\frac{1}{q} \geqslant \frac{\sin \alpha}{\sin \beta} \cdot \frac{\sin \beta}{\sin \gamma}+\frac{\sin \beta}{\sin \gamma} \\
& =\frac{\sin \alpha+\sin \beta}{\sin \gamma}=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)} .
\end{aligned}
$$

Notice that, $\alpha, \beta \leqslant \frac{\beta+\gamma}{2} \Rightarrow q<\frac{\sqrt{5}+1}{2}$.
Therefore, the required range is $\left[1, \frac{\sqrt{5}+1}{2}\right)$.
3. Given $0<\alpha \leqslant \beta \leqslant \gamma, \alpha+\beta+\gamma=\pi$. Then the range of $\min \left\{\frac{\sin \beta}{\sin \alpha}, \frac{\sin \gamma}{\sin \beta}\right\}$ is $\qquad$