13. As shown in Figure 1, given the tetrahedron $O-ABC$ with lateral edges $OA$, $OB$, and $OC$ mutually perpendicular, and $OA=1$, $OB=OC=2$, $E$ is the midpoint of $OC$.
(1) Find the distance from point $O$ to plane $ABC$;
(2) Find the angle formed by the skew lines $BE$ and $AC$;
(3) Find the size of the dihedral angle $E-AB-C$.
Three, 13. (1) As shown in Figure 4, with $O$ as the origin, $O B$, $O C$, and $O A$ as the $x$, $y$, and $z$ axes, respectively, we have:
$$
\begin{array}{l}
A(0,0,1) \text {, } \\
B(2,0,0) \text {, } \\
C(0,2,0) \\
E(0,1,0) .
\end{array}
$$

Let the normal vector of plane $A B C$ be $\boldsymbol{n}_{1}=(x, y, z)$.
Since $\boldsymbol{n}_{1} \perp \boldsymbol{A B}$, we have $\boldsymbol{n}_{1} \cdot \boldsymbol{A B}=2 x-z=0$;
Since $\boldsymbol{n}_{1} \perp \boldsymbol{A C}$, we have $\boldsymbol{n}_{1} \cdot \boldsymbol{A C}=2 y-z=0$.
Taking $\boldsymbol{n}_{1}=(1,1,2)$, the distance from point $O$ to plane $A B C$ is $d=\frac{\left|n_{1} \cdot O A\right|}{\left|n_{1}\right|}=\frac{2}{\sqrt{1+1+4}}=\frac{\sqrt{6}}{3}$.
$$
\begin{array}{l}
\text { (2) } \boldsymbol{E B}=(2,0,0)-(0,1,0)=(2,-1,0), \\
\boldsymbol{A} \boldsymbol{C}=(0,2,-1), \\
\cos \langle\boldsymbol{E B}, \boldsymbol{A} \boldsymbol{C}\rangle=\frac{-2}{\sqrt{5} \cdot \sqrt{5}}=-\frac{2}{5} .
\end{array}
$$

Thus, the angle between $B E$ and $A C$ is $\arccos \frac{2}{5}$.
(3) Let the normal vector of plane $E A B$ be $\boldsymbol{n}=(x, y, z)$.

Since $\boldsymbol{n} \perp \boldsymbol{A B}$, we have $\boldsymbol{n} \cdot \boldsymbol{A B}=2 x-z=0$;
Since $\boldsymbol{n} \perp \boldsymbol{E B}$, we have $\boldsymbol{n} \cdot \boldsymbol{E B}=2 x-y=0$.
Taking $\boldsymbol{n}=(1,2,2)$. From (1), the normal vector of plane $A B C$ is $\boldsymbol{n}_{1}=(1,1,2)$.
$$
\begin{array}{l}
\text { Then } \cos =\frac{\boldsymbol{n} \cdot \boldsymbol{n}_{1}}{|\boldsymbol{n}|\left|\boldsymbol{n}_{1}\right|} \\
=\frac{1+2+4}{\sqrt{9} \cdot \sqrt{6}}=\frac{7}{3 \sqrt{6}}=\frac{7 \sqrt{6}}{18} .
\end{array}
$$

From the diagram, the dihedral angle $E-A B-C$ is $\arccos \frac{7 \sqrt{6}}{18}$.