(1) $\frac{\sqrt{2}}{2}$ Hint: $A, B, C, D$ form a regular tetrahedron, and finding the minimum distance between $P$ and $Q$ is equivalent to finding the distance between $A B$ and $C D$, as shown in the figure.

Take the midpoints $E, F$ of $A B$ and $C D$, respectively. Then $A F \perp C D$, $B F \perp C D$, so $C D \perp$ plane $A B F$. Therefore, $E F \perp C D$. Also, $E F \perp A B$, so $E F$ is the distance between $A B$ and $C D$.
Since $A E=B E=\frac{1}{2}, A F=\frac{\sqrt{3}}{2}$, we have
$$
E F=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{2}}{2} .
$$
(1) In space, the distances between four points $A, B, C, D$ are all 1. Points $P$ and $Q$ move on line segments $AB$ and $CD$ respectively. The minimum distance between points $P$ and $Q$ is $\qquad$.