解：?$(1)$?把點(diǎn)?$A(3，$??$3)$?代入?$y=x^2+bx$?中，得?$3=9+3b$?
解得?$b=-2$?
∴二次函數(shù)的表達(dá)式為?$y=x^2-2x$?
?$(2)$?設(shè)點(diǎn)?$P $?在點(diǎn)?$Q $?的左下方，過(guò)點(diǎn)?$P $?作?$PE⊥QQ_{1}$?于點(diǎn)?$E，$?如圖①所示
∵?$PE⊥QQ_{1}，$??$QQ_{1}⊥x$?軸
∴?$PE//x$?軸
∵直線?$OA$?的表達(dá)式為?$y=x$?
∴?$∠QPE=45°$?
∴?$PE= \frac {\sqrt{2}}{2}PQ=2$?
設(shè)點(diǎn)?$P(m，$??$m)(0< m< 1)，$?則?$Q(m+2，$??$m+2)，$??$P_{1}(m，$??$\ \mathrm {m^2}-2\ \mathrm {m})，$??$Q_{1}(m+2，$??$\ \mathrm {m^2}+2\ \mathrm {m})$?
∴?$PP_{1}=3m-\ \mathrm {m^2}，$??$QQ_{1}=2-\ \mathrm {m^2}-m$?
∴?$S_{梯形PQQ_{1}P_{1}}= \frac 12(PP_{1}+QQ_{1}) · PE=-2\ \mathrm {m^2}+2m+2=-2(m-\frac 12) ^2+\frac {5}{2} $?
∴當(dāng)?$m= \frac 12$?時(shí)，?$S_{梯形PQQ_{1}P_{1}}$?取最大值，最大值為?$ \frac {5}{2} $?
?$(3)$?存在， 如圖②
①點(diǎn)?$E$?的對(duì)稱點(diǎn)為?$F，$??$EF $?與?$AM$?交于點(diǎn)?$G，$?連接?$OM、$??$MF、$??$AF、$??$OF$?
∵?$S_{△AOF}=S_{△AOM}$?
∴?$MF//OA$?
∴?$△AEG∽△MFG$?
∴?$\frac {EG}{FG}=\frac {AG}{MG}$?
∵?$EG=GF$?
∴?$AG=GM$?
∵二次函數(shù)?$y=x^2-2x$?
∴其頂點(diǎn)?$M$?的坐標(biāo)為?$(1，$??$-1)$?
∵?$A(3，$??$3)$?
∴點(diǎn)?$G(2，$??$1)$?
易求得直線?$AM$?的表達(dá)式為?$y=2x-3$?
則可求出線段?$AM$?的垂直平分線?$EF $?的表達(dá)式為?$y=- \frac {1}{2} x+2$?
由?$ \begin{cases}{y=x}\\{y=- \dfrac {1}{2} x+2}\end{cases}，$?解得?$\begin{cases}{x=\dfrac {4}{3}}\\{y= \dfrac {4}{3} }\end{cases}$?
∴點(diǎn)?$E$?的坐標(biāo)為?$(\frac 43，$??$\frac 43)$?
②設(shè)點(diǎn)?$E$?關(guān)于點(diǎn)?$A$?的對(duì)稱點(diǎn)為?$E'，$??$E'$?關(guān)于?$AM$?的對(duì)稱點(diǎn)為?$F'$?
根據(jù)對(duì)稱性可知，?$△OAF' $?與?$△AOF $?的面積相等
此時(shí)?$E'(\frac {14}{3}，$??$ \frac {14}{3} )$?
綜上所述，滿足條件的點(diǎn)?$E$?的坐標(biāo)為?$(\frac {4}{3} ，$??$\frac {4}{3} ) $?或?$(\frac {14}{3}，$??$ \frac {14}{3} )$?
