答案：
1. 2或$\sqrt{5}$或$\frac{5}{3}$ 解析：$\because$点$E$在$BC$上，且$AB = CE = \frac{1}{2}BE = 1$，$\therefore BE = 2$，$\therefore BC = BE + CE = 3$。在$Rt△ ABE$中，由勾股定理得$AE = \sqrt{AB^{2} + BE^{2}} = \sqrt{5}$，$\because$四边形$ABCD$是矩形，$\therefore CD = AB = 1$，$AD = BC = 3$，$∠ DAB = ∠ B = ∠ BCD = ∠ CDA = 90^{\circ}$，$AD// BC$。$\because$点$F$是线段$AD$上的一个动点（点$F$不与点$A$，$D$重合），$\therefore$当点$G$正好落在矩形任意一边所在的直线上时，有以下三种情况：
①当点$G$落在直线$AD$上时，则点$G$在$AD$的延长线上，如图①，由翻折的性质得$AF = GF$，$AE = GE$，$\therefore EF⊥ AG$，$\therefore ∠ DAB = ∠ B = ∠ EFA = 90^{\circ}$，$\therefore$四边形$ABEF$是矩形，$\therefore AF = BE = 2$。

②当点$G$落在直线$BC$上时，则点$G$在$BC$的延长线上，如图②，由翻折的性质得$∠ AEF = ∠ GEF$，$\because AD// BC$，$\therefore ∠ AFE = ∠ GEF$，$\therefore ∠ AEF = ∠ AFE$，$\therefore AF = AE = \sqrt{5}$。
③当点$G$落在直线$CD$上时，则点$G$在$CD$的延长线上，如图③，由翻折的性质得$EG = AE = \sqrt{5}$，$AF = FG$，则$CG = \sqrt{EG^{2} - CE^{2}} = 2$，$\therefore GD = 2 - 1 = 1$。在$Rt△ DGF$中，$FG^{2} = DG^{2} + FD^{2}$，则$AF^{2} = 1^{2} + (3 - AF)^{2}$，解得$AF = \frac{5}{3}$。

综上所述，所有满足条件的线段$AF$的长是$2$或$\sqrt{5}$或$\frac{5}{3}$。

答案：
2. (1)$\because$四边形$ABCD$是矩形，$\therefore ∠ B = 90^{\circ}$，$\therefore AC = \sqrt{AB^{2} + BC^{2}} = \sqrt{3^{2} + 4^{2}} = 5$。由矩形旋转可知$CB = CF = 4$，$\therefore AF = AC - CF = 5 - 4 = 1$。
(2)如图①，过点$C$作$CH⊥ BD$于点$H$；
在$Rt△ BDC$中，$BD = \sqrt{BC^{2} + CD^{2}} = 5$，由矩形旋转可知$CB = CF$，$\because CH⊥ BD$，$BH = FH = \frac{1}{2}BF$，$\because S_{△ BDC} = \frac{1}{2}BD· CH = \frac{1}{2}BC· CD$，$\therefore CH = \frac{12}{5}$，$\therefore BH = \sqrt{BC^{2} - CH^{2}} = \frac{16}{5}$，$\therefore BF = 2BH = \frac{32}{5}$，$\therefore DF = BF - BD = \frac{7}{5}$，$\therefore S_{△ CDF} = \frac{1}{2}DF· CH = \frac{1}{2}×\frac{7}{5}×\frac{12}{5} = \frac{42}{25}$。

(3)$BG^{2} + DF^{2}$的值为$50$。解析：如图②，连接$BD$，$GF$，易得$FG = BD = 5$，由矩形旋转可知$CB = CF$，$CD = CG$，$∠ BCF = ∠ DCG$，$\therefore ∠ CBF = ∠ CFB$，$∠ CDG = ∠ CGD$，$\therefore ∠ CBF = ∠ CDG$。$\because$四边形$ABCD$是矩形，$\therefore ∠ BCD = 90^{\circ}$，$\because ∠ CBF = ∠ CDG$，$∠ CIB = ∠ DIH$，$\therefore ∠ DHB = ∠ BCD = 90^{\circ}$，即$BF⊥ DG$。在$Rt△ BGH$中，$BG^{2} = BH^{2} + HG^{2}$，在$Rt△ DFH$中，$DF^{2} = DH^{2} + HF^{2}$，在$Rt△ BDH$中，$BD^{2} = BH^{2} + DH^{2}$，在$Rt△ FGH$中，$GF^{2} = HF^{2} + HG^{2}$，$\therefore BG^{2} + DF^{2} = BD^{2} + GF^{2} = 5^{2} + 5^{2} = 50$，即$BG^{2} + DF^{2}$的值为$50$。

答案：
3. B 解析：如图，设$PQ$，$AC$交于点$D$，过点$D$作$DE⊥ BC$于点$E$，连接$BD$，$\because$四边形$PAQC$是平行四边形，$\therefore PQ = 2PD$，$AD = CD = \frac{1}{2}AC = 2$。$\because$点$D$是$AC$的中点，为定点，$\therefore$由垂线段最短可知，当$PD⊥ BC$时，$PD$取得最小值，则$PQ$最小，即当$P$，$E$重合时，$PD$最小，$\therefore PD$的最小值为$DE$。$\because ∠ BAC = 90^{\circ}$，$AB = 3$，$AC = 4$，$\therefore BC = \sqrt{AB^{2} + AC^{2}} = 5$。$\because S_{△ ABC} = S_{△ ABD} + S_{△ BCD}$，即$\frac{1}{2}AB· AC = \frac{1}{2}AB· AD + \frac{1}{2}BC· DE$，$\therefore \frac{1}{2}×3×4 = \frac{1}{2}×3×2 + \frac{1}{2}×5× DE$，$\therefore DE = \frac{6}{5}$，$\therefore PD$的最小值为$\frac{6}{5}$，$\therefore PQ$的最小值为$2×\frac{6}{5} = \frac{12}{5}$，故选B。

答案：
4. 4 解析：$\because E$为$AB$中点，$F$为$AP$中点，$\therefore EF = \frac{1}{2}BP$，$\therefore C_{△ AEF} = AE + AF + EF = \frac{1}{2}(AB + AP + BP) = \frac{1}{2}C_{△ ABP}$，当$△ ABP$的周长最小时，$△ AEF$的周长最小，即$AP + BP$的值最小时，$△ AEF$的周长最小。
如图，作$A$关于$CD$的对称点$A'$，连接$A'B$交$CD$于$P$，连接$A'C$，$BD$，$\therefore AD = A'D = BC = 2$，$A'D// BC$，$\therefore$四边形$A'DBC$是平行四边形，$\therefore CP = DP = \frac{3}{2}$，$A'P = BP$，$\therefore AP = BP = \sqrt{CP^{2} + BC^{2}} = \sqrt{\frac{9}{4} + 4} = \frac{5}{2}$，$\therefore C_{△ AEF} = \frac{1}{2}C_{△ ABP} = \frac{1}{2}×(3 + \frac{5}{2} + \frac{5}{2}) = 4$。

答案：
5. $\frac{3}{2}$ 解析：如图，以$CE$为一边在正方形$ABCD$内作等边$△ CEH$，连接$FH$，过点$H$作$HP⊥ BC$于点$P$，过点$F$作$FT⊥ HP$于点$T$，$\because$四边形$ABCD$为正方形，且边长为$2$，$\therefore BC = 2$，$∠ B = 90^{\circ}$。$\because$点$E$为$BC$的中点，$\therefore BE = CE = 1$。$\because △ EFG$和$△ CEH$均为等边三角形，$HP⊥ BC$，$\therefore EF = EG$，$EH = EC$，$∠ FEG = ∠ CEH = 60^{\circ}$，$EP = PC = \frac{1}{2}$。$\because HP⊥ BC$，$FT⊥ HP$，$∠ B = 90^{\circ}$，$\therefore$四边形$BFTP$为矩形，$\therefore FB = TP$，$BP = FT = BE + EP = \frac{3}{2}$。$\because ∠ FEG = ∠ CEH = 60^{\circ}$，$\therefore ∠ FEG + ∠ HEG = ∠ CEH + ∠ HEG$，即$∠ FEH = ∠ CEG$。在$△ EFH$和$△ EGC$中，$\begin{cases}EF = EG,\\∠ FEH = ∠ GEC,\\EH = EC,\end{cases}$ $\therefore △ EFH≌△ EGC(\mathrm{SAS})$，$\therefore FH = CG$；$\because ∠ FTH = 90^{\circ}$，$\therefore FH≥ FT$，$\therefore$当点$T$与点$H$重合时，$FH = BP = \frac{3}{2}$为最小，即$CG$为最小，最小值为$\frac{3}{2}$。