4. $(2 \sqrt{2}, 9)$

Analysis: According to the problem, we have: $\left\{\begin{array}{c}0<a_{1}<1 \\ 1<a_{1} q<2 \\ 2<a_{1} q^{2}<3\end{array}\right.$, which means $\left\{\begin{array}{c}0<a_{1}<1 \\ \frac{1}{q}<a_{1}<\frac{2}{q} \\ \frac{2}{q}<a_{1}<\frac{3}{2}\end{array}\right.$, leading to $\sqrt{2}<q<3$, thus
$$
a_{4}=a_{3} q \in(2 \sqrt{2}, 9)
$$
4. For the geometric sequence $a_{1}, a_{2}, a_{3}, a_{4}$, it satisfies $a_{1} \in(0,1), a_{2} \in(1,2), a_{3} \in(2,3)$, then the range of $a_{4}$ is
$\qquad$