In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ $=1: 2$. Find the dihedral angle formed by the plane $B_{1} E F$ and the base $A_{1} B_{1} C_{1} D_{1}$.

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〔Solution One〕, see "Intermediate Mathematics" 1985, No. 6, P 45 ~ P 46

This method of finding the plane angle of a dihedral angle is the most basic, but when the plane angle is not obvious, it is difficult to add auxiliary lines.

〔Solution Two〕 Draw a perpendicular line $BO$ from $B$ to the plane $B_{1}EF$, with $O$ as the foot of the perpendicular.
$\because B B_{1}$ is the perpendicular line of the plane $A_{1} B_{1} C_{1} D_{1}$, according to Theorem 2, to find the dihedral angle between $\left.B_{1} E\right\}$ and the plane $A_{1} B_{1} C_{1} D_{1}$, we only need to find $\angle B_{1} B O$. And $\triangle B_{1} B O$ is a right-angled triangle. If the edge length of the cube is 1, we only need to find the length of $B O$, and the problem will be solved.
Using the volume of the tetrahedron: $V_{B-B B_{1} E F}=\frac{1}{3} S_{\triangle B_{1} E F}$
- $B O$ and $V_{F-B \mid E B}=\frac{1}{3} S_{A B_{1} B_{B}} \cdot A_{1} B_{1}$
$=\frac{1}{3}\left(\frac{1}{2} \cdot 1 \cdot \frac{1}{2}\right) \cdot 1=\frac{1}{12}$, from the equality of the two, we get
$B O=\frac{1}{4 S_{\triangle B_{1} \Sigma^{F}}}$.
Now the key is to find the area of $\triangle B_{1} E F$. Of course, we can use Heron's formula to find it, but this method is too complicated, so we have to think of another way.

Here we introduce a unique method of adding auxiliary lines, which can greatly simplify the area calculation. Draw $F G \perp B_{1} E$ from $F$, intersecting $B_{1} E$ at $G$, i.e., find the height $F G$ of $\triangle F B_{1} E$, and then draw $F H / / A_{1} B_{1}$ from $F$, intersecting $B_{1} B$ at $H$, and connect $H G$.
$$
\left.\begin{array}{r}
\because F H \perp H G \\
B_{1} E \perp F G
\end{array}\right\} \Rightarrow B_{1} E \perp H G \quad \text { (Inverse of the Three Perpendiculars }
$$

Theorem)
$\therefore \Delta B_{1} G H$ is a right-angled triangle.
Using $\triangle B_{1} G H \sim \triangle B_{1} B E$, we have $\frac{H G}{B E}$
$=\frac{H B_{1}}{B_{1} E}$, from which we get, $H G=\frac{1}{3 \sqrt{5}}$.
Thus, $F G^{2}=F H^{2}+H G^{2}=\frac{46}{45}$,
$B_{1} E=\sqrt{B B_{1}{ }^{2}+B E^{2}}=\frac{\overline{5}}{2}$,
$S_{\triangle B_{1} B P}=\frac{1}{2} \cdot B_{1} E^{\prime} \cdot F G=\frac{\sqrt{46}}{12}$.
Therefore, $B$ O $=\frac{12}{\sqrt{46}} \times \frac{1}{4}=\frac{3}{\sqrt{46}}$. Let
$\angle B_{1} B O=\alpha\left(\alpha \leqslant 90^{\circ}\right)$. Then $\operatorname{tg} \alpha=\frac{\sqrt{37}}{3}$,
$\alpha=\operatorname{tg}^{-1} \frac{\sqrt{37}}{3}$.
Thus, the dihedral angle between the plane $B_{1} E F$ and the base $A_{1} B_{1} C_{1} D_{1}$ is $\operatorname{tg}^{-1} \frac{\sqrt{37}}{3}$.
〔Analytical Method〕 Using the fact that the angle between two planes is equal to or complementary to the angle between their normal vectors, and using vector operations to solve the problem is much more convenient.
First, establish a spatial rectangular coordinate system with $B_{1}$ as the origin, the line of $B_{1} C_{1}$ as the $x$-axis, the line of $B_{1} B$ as the $z$-axis, and the line of $B_{1} A_{1}$ as the $y$-axis.
Determine the coordinates of the points from the given conditions:
$$
\begin{array}{l}
B_{1}(0,0,0), E\left(\frac{1}{2}, 0,1\right), \\
F\left(0,1, \frac{1}{3}\right) .
\end{array}
$$

The normal vector of the base $A_{1} B_{1} C_{1} D_{1}$ is $\overrightarrow{n_{1}}\{0,0,1\}$.
$$
\begin{array}{c}
\Delta B_{1} E F \text { is in the plane with normal vector } \overrightarrow{n_{2}}=\overrightarrow{B_{1} E} \\
\times \overrightarrow{B_{1} F}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
\frac{1}{2} & 0 & 1 \\
0 & 1 & \frac{1}{3}
\end{array}\right|=\left\{-1,-\frac{1}{6}, \frac{1}{2}\right\} . \\
\cos \alpha=\frac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right|\left|\overrightarrow{n_{2}}\right|}=\frac{3}{\sqrt{46}} .
\end{array}
$$
(Where $\alpha$ is the angle between the normal vectors of the two planes, choose the angle less than $90^{\circ}$)
Thus, $\operatorname{tg} \alpha=\frac{\sqrt{37}}{3}, \quad \therefore a=\operatorname{tg}^{-1} \frac{\sqrt{37}}{3}$.