（2）解：连接$AM.$
由折叠知，$AE = BE，$$AB = BM，$$\angle AEM=\angle BEM = 90^{\circ}，$
$\therefore AM = BM，$$\therefore AB = BM = AM，$
$\therefore\triangle ABM$为等边三角形，$\therefore\angle ABM = 60^{\circ}，$
$\therefore\angle CBM=\angle ABC-\angle ABM = 90^{\circ}-60^{\circ}=30^{\circ}，$
在正方形$ABCD$中，$AB = BC，$$\angle BAP=\angle C = 90^{\circ}，$
由折叠知$AB = BM，$$\angle PMB=\angle BAP = 90^{\circ}，$
$\therefore BC = BM，$$\angle BMQ=\angle C = 90^{\circ}，$
在$Rt\triangle BMQ$和$Rt\triangle BCQ$中，
$\begin{cases}BQ = BQ\\BM = BC\end{cases}，$
$\therefore Rt\triangle BMQ\cong Rt\triangle BCQ(HL)，$
$\therefore\angle MBQ=\angle CBQ，$
$\therefore\angle MBQ=\frac{1}{2}\angle CBM=\frac{1}{2}\times30^{\circ}=15^{\circ}，$
$\therefore\angle MBQ$的度数为$15^{\circ}.$
（3）解：当点$Q$在点$F$的下方时，
$\because$正方形$ABCD$中，$AD = CD = 4，$
$\therefore DQ=QF + DF=1+\frac{1}{2}CD=1 + 2 = 3，$
$\therefore CQ=CD - DQ=4 - 3 = 1.$
由（2）知$Rt\triangle BMQ\cong Rt\triangle BCQ(HL)，$$\therefore MQ = CQ = 1，$
设$AP = x，$由折叠知$MP = AP = x，$
$\therefore PQ=MP + MQ=x + 1，$$PD=AD - AP=4 - x，$
在$Rt\triangle PDQ$中，$PD^{2}+DQ^{2}=PQ^{2}，$
$\therefore(4 - x)^{2}+3^{2}=(x + 1)^{2}，$
展开得$16-8x+x^{2}+9=x^{2}+2x + 1，$
移项得$16-8x+x^{2}+9-x^{2}-2x - 1 = 0，$
合并同类项得$-10x+24 = 0，$
解得$x=\frac{12}{5}，$即$AP=\frac{12}{5}\text{ cm}；$
当点$Q$在点$F$的上方时，
则$DQ=DF - QF=\frac{1}{2}CD - 1=2 - 1 = 1，$
$\therefore CQ=CD - DQ=4 - 1 = 3，$$\therefore MQ = CQ = 3，$设$AP = MP = x，$
则$PD=AD - AP=4 - x，$$PQ=MP + MQ=x + 3，$
在$Rt\triangle PDQ$中，$PD^{2}+DQ^{2}=PQ^{2}，$
$\therefore(4 - x)^{2}+1^{2}=(x + 3)^{2}，$
展开得$16-8x+x^{2}+1=x^{2}+6x + 9，$
移项得$16-8x+x^{2}+1-x^{2}-6x - 9 = 0，$
合并同类项得$-14x + 8 = 0，$
解得$x=\frac{4}{7}，$
即$AP=\frac{4}{7}\text{ cm}.$
综上可知，$AP$的长为$\frac{12}{5}\text{ cm}$或$\frac{4}{7}\text{ cm}.$
