Let the midpoints of $AB$ and $AC$ be $K$ and $M$, respectively. It is easy to prove that the plane $ABM$ is the plane $\alpha$.
By the median length formula, we know $AM^2 = \frac{3}{2}$.
Thus, $KM = \sqrt{AM^2 + AK^2} = \sqrt{\frac{3}{2} - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{5}}{2}$.
It is also easy to see that the projection of line $PC$ onto plane $\alpha$ is line $MK$, and $CM = 1$, $KC = \frac{\sqrt{3}}{2}$. Therefore,
$$
\cos \angle KMC = \frac{KM^2 + MC^2 - KC^2}{2 \cdot KM \cdot MC} = \frac{3 \sqrt{5}}{10}.
$$

Thus, the cosine of the angle between the edge $PC$ and the plane $\alpha$ is
$$
\text{ } \frac{3 \sqrt{5}}{10} \text{. }
$$
5. In a regular tetrahedron \( P-ABC \), \( AB=1, AP=2 \). A plane \(\alpha\) passing through \( AB \) bisects its volume. Then the cosine of the angle formed by the edge \( PC \) and the plane \(\alpha\) is \(\qquad\)