5. As shown in Figure 1, in the quadrilateral pyramid $P-ABCD$, the base $ABCD$ is a square,
$\triangle PAB$ is an equilateral triangle, $O$ is the midpoint of side $AB$,
and $PO \perp$ plane $ABCD$. Then the cosine value of the dihedral angle $P-AC-D$ is $\qquad$
5. $-\frac{\sqrt{7}}{7}$.

Solution 1 As shown in Figure 4, draw $O E \perp A C$ at point $E$, and connect $P E$.

From $P O \perp$ plane $A B C D$, we know that $P E \perp A C, \angle P E O$ is the plane angle of the dihedral angle $P-A C-B$.
Let $A B=2$. Then
$$
P O=\sqrt{3}, O E=\frac{\sqrt{2}}{2}, P E=\frac{\sqrt{14}}{2} \text {. }
$$

Therefore, $\cos \angle P E O=\frac{O E}{P E}=\frac{\sqrt{7}}{7}$.
Thus, the cosine value of the dihedral angle $P-A C-D$ is $-\frac{\sqrt{7}}{7}$.
Solution 2 As shown in Figure 5, with $O$ as the origin, the rays $O B$ and $O P$ as the positive $x$-axis and $z$-axis, respectively, establish a spatial rectangular coordinate system.
Let $A B=2$. According to the problem, we have
$$
\begin{array}{l}
P(0,0, \sqrt{3}), A(-1,0,0), \\
C(1,2,0), D(-1,2,0). \\
\text { Then } \overrightarrow{A P}=(1,0, \sqrt{3}), \overrightarrow{A C}=(2,2,0),
\end{array}
$$

and $\boldsymbol{m}=(-\sqrt{3}, \sqrt{3}, 1)$ is a normal vector of plane $PAC$.

Also, $n=(0,0,1)$ is a normal vector of plane $ACD$, and the dihedral angle $P-A C-D$ is an obtuse angle, so the cosine value of the dihedral angle $P-A C-D$ is $-\frac{1}{\sqrt{7}}=-\frac{\sqrt{7}}{7}$.