5. Point $E$ is the midpoint of edge $B B_{1}$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Find the tangent of the angle between the lines $A E$ and $C A_{1}$.

- Let the edge of the cube be 1. Then $C A_{1}=\sqrt{3}$. Draw a line through point $A_{1}$ parallel to $A E$. It intersects the extension of edge $B B_{1}$ at point $F$, and

$$
\begin{gathered}
B_{1} F=\frac{1}{2}, A_{1} F=\sqrt{A_{1} B_{1}^{2}+B_{1} F^{2}}=\sqrt{1+\frac{1}{4}}=\frac{\sqrt{5}}{2} ; \\
C F=\sqrt{C B^{2}+B F^{2}}=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} .
\end{gathered}
$$

The desired angle is equal to angle $C A_{1} F$ or its adjacent angle. In triangle $C A_{1} F$

$$
\begin{gathered}
C F^{2}=C A_{1}^{2}+A_{1} F^{2}-2 C A_{1} \cdot A_{1} F \cdot \cos \angle C A_{1} F \\
\frac{13}{4}=3+\frac{5}{4}-2 \cdot \sqrt{3} \cdot \frac{\sqrt{5}}{2} \cdot \cos \angle C A_{1} F \\
\cos \angle C A_{1} F=\frac{\sqrt{15}}{15}, \quad \sin \angle C A_{1} F=\frac{\sqrt{210}}{15}, \quad \operatorname{tg} \angle C A_{1} F=\sqrt{14}
\end{gathered}
$$
Answer: $\sqrt{14}$.

Grading criteria:

Maximum points for one problem 7 points - a complete justified solution is provided,

6 points - the problem is almost solved,

5 points - paths to a complete solution have been identified,

4 points - reasoning or formulas that show significant progress have been provided,

3 points - reasoning or formulas that show some progress have been provided,

2 points - reasoning or formulas that can be considered an attempt to solve the problem have been provided,

1 point - an attempt to solve the problem has been made,

0 points - no attempt to solve, i.e., no additional constructions, no formulas, transformations, calculations aimed at solving the problem.