?$(1)$?解：設(shè)直線?$AB$?對應(yīng)的函數(shù)表達(dá)式為?$y=kx+b(k≠0)，$?
將?$A(3，$??$2)、$??$B(6，$??$4)$?代入，得?$\begin{cases}{3k+b=2}\\{6k+b=4}\end{cases}，$?解得?$\begin{cases}{k=\dfrac {2}{3}}\\{b=0}\end{cases}$?
∴直線?$AB$?對應(yīng)的函數(shù)表達(dá)式為?$y=\frac {2}{3} x$?
根據(jù)題意，得?$E(5，$??$2)、$??$N(10，$??$4)$?
設(shè)直線?$EN$?對應(yīng)的函數(shù)表達(dá)式為?$y=px+q(p≠0)$?
將點(diǎn)?$E、$??$N$?的坐標(biāo)代入，得?$\begin{cases}{5p+q=2}\\{10p+q=4}\end{cases}，$?解得?$\begin{cases}{p=\dfrac {2}{5}}\\{q=0}\end{cases}$?
∴直線?$EN$?對應(yīng)的函數(shù)表達(dá)式為?$y=\frac {2}{5} x $?
?$(2)$?∵四邊形?$ACDE$?是邊長為?$2$?的正方形，四邊形?$BGMN$?是邊長為?$4$?的正方形
∴?$∠CAE=∠GBN=90°，$??$∠ACD=∠BGM=90°，$??$∠CDE=∠GMN=90°，$??$∠AED=∠BNM=90°$?
∵?$\frac {AC}{BG}=\frac {CD}{GM}=\frac {DE}{MN}=\frac {AE}{BN}=\frac {2}{4}=\frac {1}{2}$?
∴正方形?$ACDE∽$?正方形?$BGMN$?
又∵直線?$AB、$??$EN、$??$CG、$??$DM$?都經(jīng)過點(diǎn)?$O$?
∴正方形?$ACDE$?和正方形?$BGMN$?是位似圖形
?$(3) $?如圖，正方形?$PN'B'G'、$?正方形?$PN''B''G''$?即為所求
