7. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, $E$ is the midpoint of $A B$, $F$ is the midpoint of $C C_{1}$, the cosine value of the angle formed by the skew lines $E F$ and $A C_{1}$ is . $\qquad$
Answer: $\frac{2 \sqrt{2}}{3}$.
Analysis: Let the edge length of the cube be 1, and establish a spatial rectangular coordinate system with $D A$ as the $x$-axis, $D C$ as the $y$-axis, and $D D_{1}$ as the $z$-axis. Then
$$
E\left(1, \frac{1}{2}, 0\right), F\left(0,1, \frac{1}{2}\right), A(1,0,1), \quad C_{1}(0,1,1), \overrightarrow{E F}=\left(-1, \frac{1}{2}, \frac{1}{2}\right), \overrightarrow{A C_{1}}=(-1,1,1)
$$

Therefore,
$$
\cos \theta=\frac{\overrightarrow{E F} \cdot \overrightarrow{A C_{1}}}{|\overrightarrow{E F}| \cdot\left|\overrightarrow{A C_{1}}\right|}=\frac{2 \sqrt{2}}{3}
$$