Let the foot of perpendiculars of $P$ and $Q$ on $CD$ and $AB$ be $P^\prime$ and $Q^\prime$ respectively. Then, it suffices to prove ${PP^\prime}^2={QQ^\prime}^2$ $PC sin\angle APD=PC sin\angle PCD=PP^\prime=PD sin\angle PDC=PD sin\angle BPC$ $\implies {PP^\prime}^2=PC.PD.sin\angle BPC.sin\angle APD$ Similarly, $ {QQ^\prime}^2=QB.QA.sin\angle AQD.sin\angle BQC$ $PD.sin\angle APD= $ distance of $D$ from line $AB$. $PC.sin\angle BPC=$ distance of point $B$ from line $DC$. Hence $ {PP^\prime}^2$= (distance of $D$ from line $AB$)(distance of point $B$ from line $DC$) Similarly,$ {QQ^\prime}^2$=(distance of $A$ from line $CD$)(distance of point $C$ from line $AB$) Since distance of $D$ from line $AB$ equals distance of $A$ from line $CD$ and distance of point $B$ from line $DC$ equals (distance of point $C$ from line $AB$ Hence proved.

Pretty nice problem actually. And surprisingly difficult #4 for the German olympiad in my opinion. If $AB \parallel CD$, then the claim follows immediately, so let $AB \not \parallel CD$. Consider $S = AB \cap CD$. The main observation is that it is the radical center of the three given circles. That immediately implies $|SP|=|SQ|$. Now we look at the area of $\triangle SPQ$, which is \[ \frac12 \cdot |SQ| \cdot \text{d}(P,CD) = [\triangle SPQ] = \frac12 \cdot |SP| \cdot \text{d}(Q, AB). \]That gives us $\text{d}(P,CD) = \text{d}(Q,AB)$ as desired. $\hfill \square$

Let $PX$ and $PY$ be the desired distances. Let $R=AB$$\cap$$CD$. $RP^2=RD.RC=RA.RB=RQ^2$, implying that the triangle $RPQ$ is isoceles, so the altitutdes of $P$ and $Q$ in this triangle are equal,consquently $PX=QY$.

Too easy, even for a 4. Let the foot from $P$ and $Q$ to $CD$ and $AB$ be $P_1$ and $Q_1$ respectively, and let $X=AB\cap CD$, if this intersection exists. Otherwise $AB\parallel CD$ and the result is clear. By PoP, $XQ^2=XA\cdot XB=XD\cdot XC=XP^2$, which implies $XQ=XC$. Therefore, $\triangle XPP'\cong \triangle XQQ'$ by aSA, and the result follows.