答案：
13.(1)因为从这$20$人中任意选取一人作为联络员，有$20$种等可能的结果，其中选到女生的结果有$12$种，所以$P(选到女生)=\frac{12}{20}=\frac{3}{5}$．
(2)这个游戏对甲、乙双方不公平．理由如下：画树状图如下：

由树状图可知，共有$12$种等可能的结果，其中和为偶数的结果有$4$种，和为奇数的结果有$8$种，所以$P(甲参加)=\frac{4}{12}=\frac{1}{3}$，$P(乙参加)=\frac{8}{12}=\frac{2}{3}$．因为$\frac{1}{3}\neq\frac{2}{3}$，所以这个游戏对甲、乙双方不公平．

答案：14.(1)因为四边形$ABCD$为矩形，$B(2,0)$，$BC = 2\sqrt{3}$，所以$C(2,2\sqrt{3})$．因为所求抛物线以直线$x = 4$为对称轴，所以可设该抛物线的函数表达式为$y = a(x - 4)^{2}+h$．因为该抛物线经过点$O(0,0)$，$C(2,2\sqrt{3})$，所以$\begin{cases}16a + h = 0, \\ 4a + h = 2\sqrt{3},\end{cases}$解得$\begin{cases}a = -\frac{\sqrt{3}}{6}, \\ h = \frac{8\sqrt{3}}{3},\end{cases}$所以该抛物线的函数表达式为$y = -\frac{\sqrt{3}}{6}(x - 4)^{2}+\frac{8\sqrt{3}}{3}$，即$y = -\frac{\sqrt{3}}{6}x^{2}+\frac{4\sqrt{3}}{3}x$．
(2)在$y = -\frac{\sqrt{3}}{6}x^{2}+\frac{4\sqrt{3}}{3}x$中，令$y = 0$，得$-\frac{\sqrt{3}}{6}x^{2}+\frac{4\sqrt{3}}{3}x = 0$，解得$x_1 = 0$，$x_2 = 8$，所以$N(8,0)$．设$M(x,y)(2＜x＜8)$．过点$M$作$MG⊥ x$轴于点$G$，则$S_{\triangle CMN} = S_{\triangle MNG}+S_{四边形MGBC} - S_{\triangle CBN} =\frac{1}{2}(8 - x)y+\frac{1}{2}(y + 2\sqrt{3})(x - 2)-\frac{1}{2}×(8 - 2)×2\sqrt{3} = 3y+\sqrt{3}x - 8\sqrt{3} = 3(-\frac{\sqrt{3}}{6}x^{2}+ \frac{4\sqrt{3}}{3}x)+\sqrt{3}x - 8\sqrt{3} = -\frac{\sqrt{3}}{2}x^{2}+5\sqrt{3}x - 8\sqrt{3} = -\frac{\sqrt{3}}{2}(x - 5)^{2}+\frac{9\sqrt{3}}{2}$．因为$-\frac{\sqrt{3}}{2}＜0$，$2＜x＜8$，所以当$x = 5$时，$S_{\triangle CMN}$取最大值$\frac{9\sqrt{3}}{2}$．故$\triangle CMN$面积的最大值为$\frac{9\sqrt{3}}{2}$．

答案：
15.(1)由旋转的性质，得$DF = DE$，$\angle EDF = 90^{\circ}$，所以$\angle CDE + \angle CDF = 90^{\circ}$．因为四边形$ABCD$是正方形，所以$AD = CD$，$\angle ADC = 90^{\circ}$，所以$\angle CDE + \angle ADE = 90^{\circ}$，所以$\angle ADE = \angle CDF$．在$\triangle ADE$和$\triangle CDF$中，$\begin{cases}AD = CD, \\ \angle ADE = \angle CDF, \\ DE = DF,\end{cases}$所以$\triangle ADE\cong \triangle CDF$，所以$AE = CF$．
(2)如图①，过点$F$作$FG⊥ BC$，交$BC$的延长线于点$G$，过点$E$作$EK⊥ AB$于点$K$，则$\angle G = \angle AKE = 90^{\circ}$．因为$\angle B = 90^{\circ}$，所以$\angle AKE = \angle B$，所以$EK// BC$，所以$\triangle AEK∼\triangle AOB$，所以$\frac{AE}{AO}=\frac{AK}{AB}=\frac{EK}{OB}$．因为$O$是$BC$的中点，$BC = AB = 2\sqrt{5}$，所以$OB = OC =\frac{1}{2}BC =\sqrt{5}$，所以$AO =\sqrt{AB^{2}+OB^{2}} = 5$．因为$OE = 2$，所以$AE = AO - OE = 3$，所以$\frac{3}{5}=\frac{AK}{2\sqrt{5}}=\frac{EK}{\sqrt{5}}$，所以$AK = \frac{6\sqrt{5}}{5}$，$EK =\frac{3\sqrt{5}}{5}$．因为$\triangle ADE\cong\triangle CDF$，所以$\angle DAE = \angle DCF$．因为$\angle BAD = \angle DCG = 90^{\circ}$，所以$\angle EAK + \angle DAE = 90^{\circ}$，$\angle FCG + \angle DCF = 90^{\circ}$，所以$\angle FCG = \angle EAK$．在$\triangle CFG$和$\triangle AEK$中，$\begin{cases}\angle G = \angle AKE, \\ \angle FCG = \angle EAK,\end{cases}$所以$\triangle CFG\cong\triangle AEK$，所以$\begin{cases}CF = AE, \\ FG = EK =\frac{3\sqrt{5}}{5}, \\ CG = AK =\frac{6\sqrt{5}}{5},\end{cases}$所以$OG = OC + CG =\frac{11\sqrt{5}}{5}$，所以$OF =\sqrt{FG^{2}+OG^{2}} =\sqrt{26}$．

(3)如图②，延长$BA$至点$P$，使$PA = OC =\sqrt{5}$，连接$PE$，$PO$．因为$\angle DAE = \angle DCF$，$\angle DAP = \angle DCB = 90^{\circ}$，所以$\angle DAE + \angle DAP = \angle DCF + \angle DCB$，所以$\angle PAE = \angle OCF$．在$\triangle PAE$和$\triangle OCF$中，$\begin{cases}PA = OC, \\ \angle PAE = \angle OCF, \\ AE = CF,\end{cases}$所以$\triangle PAE\cong \triangle OCF$，所以$PE = OF$，所以当$PE$的长最小时，$OF$的长最小．因为$OE = 2$，所以点$E$在以点$O$为圆心，$2$为半径的半圆上运动，所以当$P$，$E$，$O$三点共线时，$PE$的长最小，且$PE = OP - OE$．因为$AB = 2\sqrt{5}$，所以$BP = PA + AB = 3\sqrt{5}$．因为$\angle B = 90^{\circ}$，$OB =\sqrt{5}$，所以$OP =\sqrt{BP^{2}+OB^{2}} = 5\sqrt{2}$，所以$PE$长的最小值为$5\sqrt{2}-2$，即$OF$长的最小值为$5\sqrt{2}-2$．