10. As shown in Figure 2, in the regular tetrahedron $A B C D$, $A E=\frac{1}{4} A B$, $C F=\frac{1}{4} C D$. Then the angle between the lines $D E$ and $B F$ is ( ).
(A) $\arccos \frac{4}{13}$
(B) $\arccos \frac{3}{13}$
(C) $\pi-\arccos \frac{4}{13}$
(D) $\pi-\arccos \frac{3}{13}$
10.A.

Let the edge length of a regular tetrahedron be $a$, then
$$
\begin{array}{l}
A E=C F=\frac{1}{4} a, \\
D E^{2}=a^{2}+\left(\frac{1}{4} a\right)^{2}-\frac{1}{2} a^{2} \cos 60^{\circ}=\frac{13}{16} a^{2} .
\end{array}
$$

Thus, $D E=\frac{\sqrt{13}}{4} a$.
Similarly, $B F=\frac{\sqrt{13}}{4} a$.
As shown in Figure 6, construct $\square B C D G$, then $D G / / B C, B G / / C D$, and
$$
B G=D G=a .
$$

Since $A D \perp B C$, it follows that $A D \perp D G$. Therefore,
$$
A G=\sqrt{2} a \text {. }
$$

Take a point $H$ on $G B$ such that $G H=\frac{1}{4} a$, then
$$
G H=C F=A E \text {. }
$$

Since $B A=B G$, it follows that $E H / / A G$, and $E H=\frac{3 \sqrt{2}}{4} a$.
Since $B H / / F D$, and $B H=F D=\frac{3}{4} a$, it follows that,
$$
D H / / B F \text {, and } D H=B F=\frac{\sqrt{13}}{4} a \text {. }
$$

The angle formed by the lines $D E$ and $D H$ is the same as the angle formed by the lines $D E$ and $B F$. Therefore,
$$
\cos \angle E D H=\frac{\left(\frac{\sqrt{13}}{4} a\right)^{2}+\left(\frac{\sqrt{13}}{4} a\right)^{2}-\left(\frac{3 \sqrt{2}}{4} a\right)^{2}}{2 \times \frac{\sqrt{13}}{4} a \times \frac{\sqrt{13}}{4} a}=\frac{4}{13} .
$$