13. As shown in the figure, in a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ and $F$ are the midpoints of edges $A B$ and $B C$ respectively.
(1) Find the size of the dihedral angle $B-F B_{1}-E$.
(2) Find the distance from point $D$ to the plane $B_{1} E F$.
(3) Can a point $M$ be found on the edge $D D_{1}$ such that $B M \perp$ plane $E F B_{1}$? If so, determine the position of point $M$; if not, explain why.
13. (I) Draw $B H \perp B_{1} F$, with the foot of the perpendicular at $H$, and connect $E H$. By the properties of a cube, we know that $E B \perp$ plane $B B_{1} F$, so $B H$ is the projection of $E H$ in plane $B B_{1} F$.
By the theorem of three perpendiculars, we know that $E H \perp B_{1} F$, thus $\angle E H B$ is the plane angle of the dihedral angle $E-B_{1} F-B$.
In the right triangle $\triangle E B H$, given that $B B_{1}=2 B F=a$, we have $B H=\frac{B F \cdot B B_{1}}{B_{1} F}=\frac{a}{\sqrt{5}}, \tan \angle E H B=\frac{E B}{B H}=\frac{\sqrt{5}}{2}$, so $\angle E H B=\arctan \frac{\sqrt{5}}{2}$, which means the size of the dihedral angle $B-B_{1} F-E$ is $\arctan \frac{\sqrt{5}}{2}$.
( II ) Since $B_{1} E=B_{1} F=D E=D F, E F$ is a common side, thus $\triangle D E F \cong \triangle B_{1} E F, S_{\triangle B_{1} \mathrm{EF}}=S_{\triangle D E F}$.
Let the distance from point $D$ to plane $B_{1} E F$ be $h$, by $V_{B_{1}-D E F}=V_{D-B_{1} E F}$, we get $\frac{1}{3} S_{\triangle D E F} \cdot B B_{1}=\frac{1}{3} S_{\triangle B_{1} E F} \cdot h$.
Thus $h=B B_{1}=a$, meaning the distance from point $D$ to plane $B_{1} E F$ is $a$.
( III) Let $E F$ intersect $B D$ at $G$, and connect $B_{1} G$.
Since $E F \perp B D, E F \perp B B_{1}$, we have $E F \perp$ plane $B B_{1} D_{1} D$, and plane $B_{1} E F \perp$ plane $B B_{1} D_{1} D$. In plane $B D_{1}$, draw $B K \perp$ $B_{1} G$ at $K$, and extend it to intersect $D D_{1}$ at $M$. By the property of two perpendicular planes, we know that $B M \perp$ plane $B_{1} E F$, meaning there exists a point $M$ on $D D_{1}$ that satisfies the condition.

In plane $B D_{1}$, since $\triangle B_{1} B G \backsim \triangle B D M$, we have $\frac{B_{1} B}{B G}=\frac{B D}{D M}$. Given $B G=\frac{\sqrt{2}}{4} a, B D=\sqrt{2} a, B B_{1}=a$, we get $D M=$ $\frac{a}{2}, M$ is the midpoint of $D D_{1}$.