(4) $\frac{\sqrt{31}}{2}$ Hint: Since $BD$ is the angle bisector of $\angle ABC$, $AB=2, BC=4$, so $\frac{AD}{DC}=\frac{AB}{BC}=\frac{1}{2}$.
Let $AD=x$, then $DC=2x$.
Thus,
$$
\begin{array}{l}
AD^{2}=AB^{2}+BD^{2}-2 AB \cdot BD \cdot \cos \frac{B}{2}, \\
DC^{2}=BC^{2}+BD^{2}-2 BC \cdot BD \cdot \cos \frac{B}{2} .
\end{array}
$$

That is,
$$
\begin{array}{r}
x^{2}=4+6-2 \cdot 2 \cdot \sqrt{6} \cos \frac{B}{2}, \\
4 x^{2}=16+6-2 \cdot 4 \cdot \sqrt{6} \cos \frac{B}{2} .
\end{array}
$$

Solving the two equations simultaneously, we get $x=1 \Rightarrow AD=1, DC=2$.
Therefore,
$$
\cos A=\frac{2^{2}+1^{2}-6}{2 \cdot 2 \cdot 1}=-\frac{1}{4} .
$$

Hence,
$$
\begin{aligned}
BE & =\sqrt{AB^{2}+AE^{2}-2 AB \cdot AE \cdot \cos A} \\
& =\sqrt{4+\frac{9}{4}+2 \cdot 2 \cdot \frac{3}{2} \cdot \frac{1}{4}} \\
& =\frac{\sqrt{31}}{2} .
\end{aligned}
$$
(4) In $\triangle A B C$, it is known that $A B=2, B C=4, \angle B$'s bisector $B D=\sqrt{6}$, then the median $B E$ on side $A C$ is $\qquad$.