13. As shown in the figure, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, point $E$ is the midpoint of edge $B C$, and point $F$ is a moving point on edge $C D$. If $D_{1} E \perp$ plane $A B_{1} F$, find the size of the dihedral angle $C_{1}-E F-A$ (express the result using inverse trigonometric functions).
13. Connect $A_{1} B$, then $A_{1} B$ is the projection of $D_{1} E$ in the plane $A B B_{1} A_{1}$,

Since $A B_{1} \perp A_{1} B$, it follows that $D_{1} E \perp A B_{1}$.
Thus $D_{1} E \perp$ plane $A B_{1} F \Leftrightarrow D_{1} E \perp A F$.
Connect $D E$, then $D E$ is the projection of $D_{1} E$ in the base plane $A B C D$,
so $D_{1} E \perp A F \Leftrightarrow D E \perp A F$.
Since $A B C D$ is a square and $E$ is the midpoint of $B C$,
it follows that $D E \perp A F$ if and only if $F$ is the midpoint of $C D$,
i.e., when point $F$ is the midpoint of $C D$, $D_{1} F \perp$ plane $A B_{1} F$.
Connect $E F$, then $E F \parallel B D$.
Connect $A C$, and let $A C$ intersect $E F$ at point $H$, then $C H \perp E F$.
Connect $C_{1} H$, then $C H$ is the projection of $C_{1} H$ in the base plane $A B C D$,
so $C_{1} H \perp E F$, hence $\angle C_{1} H C$ is the plane angle of the dihedral angle $C_{1}-E F-C$.
In the right triangle $\triangle C_{1} C H$,
since $C_{1} C=1, C H=\frac{1}{4} A C=\frac{\sqrt{2}}{4}$, it follows that $\tan \angle C_{1} H C=\frac{C_{1} C}{C H}=2 \sqrt{2}$.
Thus $\angle C_{1} H C=\arctan 2 \sqrt{2}$, and therefore $\angle A H C_{1}=\pi-\arctan 2 \sqrt{2}$.
Hence, the size of the dihedral angle $C_{1}-E F-A$ is $\pi-\arctan 2 \sqrt{2}$.