2. In a regular tetrahedron $ABCD$, let
$$
\overrightarrow{AE}=\frac{1}{4} \overrightarrow{AB}, \overrightarrow{CF}=\frac{1}{4} \overrightarrow{CD} \text {, }
$$
$\vec{U} \overrightarrow{DE}$ and $\overrightarrow{BF}$ form an angle $\theta$. Then $\cos \theta=$
2. $-\frac{4}{13}$.

Let the edge length of the regular tetrahedron be 4. Then
$$
\begin{array}{l}
\overrightarrow{B F} \cdot \overrightarrow{D E}=(\overrightarrow{B C}+\overrightarrow{C F}) \cdot(\overrightarrow{D A}+\overrightarrow{A E}) \\
=\overrightarrow{C F} \cdot \overrightarrow{D A}+\overrightarrow{B C} \cdot \overrightarrow{A E}=2 \times 4 \cos \frac{2 \pi}{3}=-4 .
\end{array}
$$

And $|\overrightarrow{B F}|=|\overrightarrow{D E}|$
$$
\begin{array}{l}
=\sqrt{\overrightarrow{B C}^{2}+\overrightarrow{C F}^{2}-2|\overrightarrow{B C}||\overrightarrow{C F}| \cos \frac{\pi}{3}} \\
=\sqrt{13},
\end{array}
$$

Then $\cos \theta=\frac{\overrightarrow{B F} \cdot \overrightarrow{D E}}{|\overrightarrow{B F}| \cdot|\overrightarrow{D E}|}=-\frac{4}{13}$.