5. Given a regular tetrahedron $S-ABC$ with the base being an equilateral triangle of side length 1, and the side edges being 2. A section through $AB$ divides the volume of the tetrahedron into two equal parts. Then the cosine of the plane angle of the dihedral angle formed by the section and the base is $\qquad$
5. $\frac{2 \sqrt{15}}{15}$.

Let the section intersect edge $S C$ at point $D$.
From the conditions, we know that $D$ is the midpoint of edge $S C$. Take the midpoint $E$ of $A B$. Thus, $\angle D E C$ is the plane angle of the dihedral angle formed by the section and the base.
In $\triangle S E C$, it is easy to know that
$E C=\frac{\sqrt{3}}{2}, S C=2, S E=\frac{\sqrt{15}}{2}$
$\Rightarrow$ Median $D E=\frac{\sqrt{5}}{2}$.
Therefore, in $\triangle D E C$, by the cosine rule we get
$$
\cos \angle D E C=\frac{2 \sqrt{15}}{15} .
$$