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 $A A_{1}$ such that $A_{1} F: F A=1: 2$. Find the dihedral angle formed by the plane $B_{1} E F$ and the base $A_{1} B_{1} C_{1} D_{1}$.
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 $D_{1} A_{1}$.
$$
\begin{array}{lc}
\because & \text { plane } A A_{1} D_{1} D / / \text { plane } B B_{1} C_{1} C, \\
\therefore & F G / / B_{1} E \\
\text { Also } & F A / / B_{1} B, \\
\therefore & \angle A F G=\angle B B_{1} E .
\end{array}
$$

Therefore,
$$
\begin{aligned}
\operatorname{tg} \angle A F G & =B E: B B_{1} \\
& =\frac{1}{2} .
\end{aligned}
$$

Assume the edge length of the cube is 6.
From
$$
A_{1} F: F A=1: 2
$$

we get
$$
A_{1} F=2, \quad F A=4, \quad H A_{1}=A_{1} F \cdot \operatorname{tg} \angle H F A_{1}=1 .
$$
$$
H B_{1}=\sqrt{A_{1} H^{2}+A_{1} B_{1}^{2}}=\sqrt{37} \text {. }
$$

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.
$$
\begin{array}{cc}
\because \quad & A_{1} K=\frac{A_{1} H \cdot A_{1} B_{1}}{H B_{1}}=\frac{6}{\sqrt{37}}, \\
\therefore \quad & \operatorname{tg} \angle F K A_{1}=\frac{F A_{1}}{A_{1} K}=\frac{\sqrt{37}}{3}, \\
& \angle F K A_{1}=\operatorname{arctg} \frac{\sqrt{37}}{3} .
\end{array}
$$