14. As shown in the figure, an equilateral triangle $ABC$ with side length 4 is folded along a line $EF$ parallel to side $BC$, such that plane $AEF \perp$ plane $BCFE$, and $O$ is the midpoint of $EF$.
(1) Find the cosine value of the dihedral angle $F-AE-B$;
(2) If $BE \perp$ plane $AOC$, find the length of the fold line $EF$.
(1) Establish a rectangular coordinate system as shown in the figure. Let $E F=2 a$, then $E(a, 0,0), F(-a, 0,0), A(0,0, \sqrt{3} a)$, $B(2,2 \sqrt{3}-\sqrt{3} a, 0)$. Therefore, $\overrightarrow{A E}=(a, 0,-\sqrt{3} a), \overrightarrow{E B}=(2-a, \sqrt{3}(2-a), 0)$, the normal vector of plane $A E F$ is $\boldsymbol{m}=(0,1,0)$, the normal vector of plane $A E B$ is $\boldsymbol{n}=\left(x_{1}, y_{1}, z_{1}\right)$, we have
$$
\left\{\begin{array}{l}
a x_{1}-\sqrt{3} a z_{1}=0, \\
(2-a) x_{1}+\sqrt{3}(2-a) y_{1}=0
\end{array} \Rightarrow \boldsymbol{n}=(\sqrt{3},-1,1) \Rightarrow \cos \langle\boldsymbol{m}, \boldsymbol{n}\rangle=-\frac{1}{\sqrt{5}}\right. \text {, }
$$

Thus, the cosine value of the dihedral angle $F-A E-B$ is $-\frac{\sqrt{5}}{5}$.
(2) Since $C(-2,2 \sqrt{3}-\sqrt{3} a, 0)$, the normal vector of plane $A O C$ is $\boldsymbol{p}=\left(x_{2}, y_{2}, z_{2}\right)$, we have
$$
\left\{\begin{array}{l}
\sqrt{3} a z_{1}=0, \\
-2 x_{1}+\sqrt{3}(2-a) y_{2}=0
\end{array} \Rightarrow \boldsymbol{n}=(\sqrt{3}(2-a), 2,0)\right. \text {. }
$$

Thus, $\overrightarrow{E B} / / \boldsymbol{p} \Rightarrow 3(2-a)^{2}=2(2-a) \Rightarrow a=\frac{4}{3}$.
In summary, when $B E \perp$ plane $A O C$, the length of the fold $E F$ is $\frac{8}{3}$.