答案：
13.(1)如图所示：

(2)设声音在空气中传播的速度$v(m/s)$与气温$t(^{\circ}C)$之间的函数表达式为$v = kt + b$.把点$(0,331)$，$(20,343)$分别代入$v = kt + b$，得$\begin{cases}b = 331,\\20k + b = 343,\end{cases}$解得$\begin{cases}k = \frac{3}{5},\\b = 331,\end{cases}$所以$v = \frac{3}{5}t + 331$.故声音在空气中传播的速度$v(m/s)$与气温$t(^{\circ}C)$之间的函数表达式为$v = \frac{3}{5}t + 331$.
(3)在$v = \frac{3}{5}t + 331$中，令$t = 23$，得$v = \frac{3}{5}×23 + 331 = 344.8$.因为$344.8×5 = 1724(m)$，所以此人与烟花燃放地相距$1724m$.

答案：14.(1)因为四边形$ABCD$为矩形，$BC = 8$，所以$\angle A = 90^{\circ}$，$AD = BC = 8$.由折叠的性质，得$\angle BA^{\prime}E = \angle A = 90^{\circ}$，$\angle A^{\prime}BE = \angle ABE$，所以$\angle ABA^{\prime}=2\angle ABE$.因为$\angle A+\angle ABA^{\prime}+\angle BA^{\prime}E+\angle AEA^{\prime}=360^{\circ}$，所以$\angle ABA^{\prime}+\angle AEA^{\prime}=180^{\circ}$.又$\angle DEA^{\prime}+\angle AEA^{\prime}=180^{\circ}$，所以$\angle DEA^{\prime}=\angle ABA^{\prime}$，所以$\angle DEA^{\prime}=2\angle ABE$.
(2)由折叠的性质，得$A^{\prime}B = AB = 6$，$A^{\prime}E = AE$.设$A^{\prime}E = AE = x$.因为$AD = BC = 8$，所以$DE = AD - AE = 8 - x$.因为$\angle A = 90^{\circ}$，所以$BD = \sqrt{AB^{2}+AD^{2}} = 10$，所以$A^{\prime}D = BD - A^{\prime}B = 4$.因为$\angle BA^{\prime}E = 90^{\circ}$，所以$\angle DA^{\prime}E = 180^{\circ}-\angle BA^{\prime}E = 90^{\circ}$，所以$A^{\prime}D^{2}+A^{\prime}E^{2}=DE^{2}$，所以$4^{2}+x^{2}=(8 - x)^{2}$，解得$x = 3$，所以$AE = 3$，所以$\tan\angle ABE = \frac{AE}{AB}=\frac{1}{2}$.

答案：
15.(1)在$y = -\frac{1}{2}x^{2}-\frac{3}{2}x + 2$中，令$x = 2$，得$y = - 3$，所以$A(2,-3)$.把点$A(2,-3)$，$C(0,-3)$分别代入$y = x^{2}+bx + c$，得$\begin{cases}4 + 2b + c = - 3,\\c = - 3,\end{cases}$解得$\begin{cases}b = - 2,\\c = - 3,\end{cases}$所以抛物线$L_{1}$的函数表达式为$y = x^{2}-2x - 3$.
(2)设$P(m,m^{2}-2m - 3)$，$Q(n,-\frac{1}{2}n^{2}-\frac{3}{2}n + 2)$.又$A(2,-3)$，$C(0,-3)$，所以当以$A$，$C$，$P$，$Q$四点为顶点的四边形恰为平行四边形时，分类讨论如下：①若$AC$为平行四边形的对角线，则$\begin{cases}m + n = 2,\\m^{2}-2m - 3+(-\frac{1}{2}n^{2}-\frac{3}{2}n + 2)= - 3+(-3),\end{cases}$解得$\begin{cases}m = - 3,\ = 5\end{cases}$或$\begin{cases}m = 0,\ = 2\end{cases}$（不合题意，舍去），则$m^{2}-2m - 3 = 12$，所以$P(-3,12)$；②若$AP$为平行四边形的对角线，则$\begin{cases}m + 2 = n,\\m^{2}-2m - 3+(-3)=-\frac{1}{2}n^{2}-\frac{3}{2}n + 2+(-3),\end{cases}$解得$\begin{cases}m = - 1,\ = 1\end{cases}$或$\begin{cases}m = 0,\ = 2\end{cases}$（不合题意，舍去），则$m^{2}-2m - 3 = 0$，所以$P(-1,0)$；③若$AQ$为平行四边形的对角线，则$\begin{cases}n + 2 = m,\\-\frac{1}{2}n^{2}-\frac{3}{2}n + 2+(-3)=m^{2}-2m - 3+(-3),\end{cases}$解得$\begin{cases}m = 3,\ = 1\end{cases}$或$\begin{cases}m = -\frac{4}{3},\ = -\frac{10}{3}\end{cases}$当$m = 3$时，$m^{2}-2m - 3 = 0$；当$m = -\frac{4}{3}$时，$m^{2}-2m - 3 = \frac{13}{9}$.所以$P(3,0)$或$P(-\frac{4}{3},\frac{13}{9})$.综上所述，点$P$的坐标为$(-3,12)$或$(-1,0)$或$(3,0)$或$(-\frac{4}{3},\frac{13}{9})$.
(3)如图，当点$P$在$y$轴左侧时，抛物线$L_{1}$上不存在点$R$，使得$CA$平分$\angle PCR$；当点$P$在$y$轴右侧时，不妨设点$P$在$CA$的上方，点$R$在$CA$的下方.过$P$，$R$两点作$y$轴的垂线，垂足分别为$S$，$T$，过点$P$作$PH⊥ TR$交$TR$的延长线于点$H$，则$\angle PSC = \angle RTC = 90^{\circ}$.因为$A(2,-3)$，$C(0,-3)$，所以$AC// x$轴，所以$AC⊥ y$轴，所以$\angle ACS = \angle ACT = 90^{\circ}$.因为$CA$平分$\angle PCR$，所以$\angle PCA = \angle RCA$.又$\angle PCA+\angle PCS = \angle ACS$，$\angle RCA + \angle RCT = \angle ACT$，所以$\angle PCS = \angle RCT$，所以$\triangle PSC∼\triangle RTC$，所以$\frac{PS}{CS}=\frac{RT}{CT}$.设$P(x_{1},x_{1}^{2}-2x_{1}-3)$，$R(x_{2},x_{2}^{2}-2x_{2}-3)$，则$\frac{x_{1}}{x_{1}^{2}-2x_{1}-3-(-3)}=\frac{x_{2}}{-3-(x_{2}^{2}-2x_{2}-3)}$.整理，得$x_{1}+x_{2} = 4$.在$Rt\triangle PRH$中，$\tan\angle PRH = \frac{PH}{RH}=\frac{x_{1}^{2}-2x_{1}-3-(x_{2}^{2}-2x_{2}-3)}{x_{1}-x_{2}}=x_{1}+x_{2}-2 = 2$.过点$Q$作$QK⊥ x$轴于点$K$.设$Q(t,-\frac{1}{2}t^{2}-\frac{3}{2}t + 2)$.若$OQ// PR$，则$\angle QOK = \angle PRH$，所以$-\frac{1}{2}t^{2}-\frac{3}{2}t + 2 = \tan\angle PRH = 2$，所以$-\frac{1}{2}t^{2}-\frac{3}{2}t + 2 = - 7\pm\sqrt{65}$，解得$t = \frac{-7\pm\sqrt{65}}{2}$，则$-\frac{1}{2}t^{2}-\frac{3}{2}t + 2 = - 7\pm\sqrt{65}$，所以点$Q$的坐标为$(\frac{-7+\sqrt{65}}{2},-7+\sqrt{65})$或$(\frac{-7-\sqrt{65}}{2},-7-\sqrt{65})$.