II. As shown in the figure, in the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $B C$, and $F$ is on $A A_{1}$, with $A_{1} F: F A=1: 2$. Find the dihedral angle formed by the plane $B_{1} E F$ and the base $A_{1} B_{1} C_{1} D_{1}$.
$$
\arctan \frac{\sqrt{37}}{3}
$$

II. 【Analysis and Solution】Let the intersection line of planes $B_{1} E F$ and $A A_{1} D_{1} D$ be $G F H$, where $G$ is on $A D$ and $H$ is on the extension of $D_{1} A_{1}$.
$\because$ Plane $A A_{1} D_{1} D \parallel$ Plane $B B_{1} C_{1} C$,
$\therefore F G \parallel B_{1} E$.
Also, $F A \parallel B_{1} B$,
$$
\therefore \angle A F G = \angle B B_{1} E.
$$

Therefore, $\tan \angle A F G = B E : B B_{1}$.
Assume the edge length of the cube is 6,
Given $A_{1} F : F A = 1 : 2$,
we get $A_{1} F = 2, F A = 4, H A_{1} = A_{1} F \cdot \tan \angle H F A_{1} = 1$,
$$
H B_{1} = \sqrt{A_{1} H^{2} + A_{1} B_{1}^{2}} = \sqrt{37}.
$$

Construct $A_{1} K \perp H B_{1}$ (where $K$ is the foot of the perpendicular), and connect $F K$, then $F K \perp H B_{1}$, and $\angle F K A_{1}$ is the plane angle of the dihedral angle we are looking for.
$$
\begin{array}{l}
\because A_{1} K = \frac{A_{1} H \cdot A_{1} B_{1}}{H B_{1}} = \frac{6}{\sqrt{37}} \\
\therefore \tan \angle F K A_{1} = \frac{F A_{1}}{A_{1} K} = \frac{\sqrt{37}}{3}, \angle F K A_{1} = \arctan \frac{\sqrt{37}}{3}.
\end{array}
$$