Exercise 1: You don't need a similar solution: pass QG⊥CD after point Q, and point G is the vertical foot, and connect QE.

Let PQ=x, from the properties of folding and rectangle,

EQ=PQ=x，QG=PD=3，EG=x-2，

In Rt△EGQ, it is obtained by Pythagorean theorem

EG 2 +GQ 2 =EQ 2, that is, (x-2) 2+3 2 = x 2.

Solution: x= 13/4, that is, PQ= 13/4. /4.

So, the answer is: 13/4. Method 2: Make auxiliary line PE, and the dotted line should be the middle vertical line L of PE (indicated by dotted line, intersecting PE at G point and PF at Q point). Sine theorem shows that PE= 13 and PG = PE/2 under the root sign. Because the triangle PDE is similar to QGP, PG/PQ = 2/.