1. Four points $A, B, C, D$ in space are pairwise 1 unit apart, and points $P, Q$ move on line segments $AB, CD$ respectively. The minimum distance between point $P$ and $Q$ is
$-1 . \frac{\sqrt{2}}{2}$. 
From the problem, we know that the tetrahedron $ABCD$ is a regular tetrahedron. Therefore, finding the minimum distance between points $P$ and $Q$ is equivalent to finding the distance between $AB$ and $CD$, as shown in Figure 1.

Take the midpoints $E$ and $F$ of $AB$ and $CD$, respectively. Then $AF \perp CD, BF \perp CD$ $\Rightarrow CD \perp$ plane $ABF \Rightarrow EF \perp CD$. Also, $EF \perp AB$, so $EF$ is the distance between $AB$ and $CD$. Hence, $AE = BE = \frac{1}{2}, AF = \frac{\sqrt{3}}{2}$ $\Rightarrow EF = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2} - \left(\frac{1}{2}\right)^{2}} = \frac{\sqrt{2}}{2}$.