Answer: $[\sqrt{5}, 2 \sqrt{5}]$.
Solution: Without loss of generality, let $\vec{e}=(1,0)$. Since $\vec{a} \cdot \vec{e}=2$, we can set $\vec{a}=(2, s)$, then for any real number $t$, we have
$$
4+s^{2}=|\vec{a}|^{2} \leq 5|\vec{a}+t \vec{e}|=5 \sqrt{(2+t)^{2}+s^{2}},
$$

which is equivalent to $4+s^{2} \leq 5|s|$, solving this yields $|s| \in[1,4]$, i.e., $s^{2} \in[1,16]$.
Thus, $|\vec{a}|=\sqrt{4+s^{2}} \in[\sqrt{5}, 2 \sqrt{5}]$.
3. In the Cartesian coordinate system, $\vec{e}$ is a unit vector, and vector $\vec{a}$ satisfies $\vec{a} \cdot \vec{e}=2$, and $|\vec{a}|^{2} \leq 5|\vec{a}+t \vec{e}|$ for any real number $t$, then the range of $|\vec{a}|$ is $\qquad$ .