解：?$(1)$?∵拋物線?$y=ax^2-2ax+c $?過點(diǎn)?$C(2，$??$3)、$??$E(-2，$??$0)$?
∴?$\begin{cases}{4a-4a+c=3}\\{4a+4a+c=0}\end{cases}，$?解得?$\begin{cases}{a=- \dfrac {3}{8}}\\{c=3}\end{cases}$?
∴拋物線的表達(dá)式為?$y=- \frac {3}{8} x^2+ \frac {3}{4} x+3$?
當(dāng)?$y=0$?時(shí)，?$- \frac {3}{8} x^2+ \frac {3}{4} x+3=0$?
解得?$x_{1}=-2($?舍去)，?$x_{2}=4$?
∴?$F(4，$??$0) $?
?$(2)$?設(shè)直線?$CE$?的表達(dá)式為?$y=kx+b$?
∵直線過點(diǎn)?$C(2，$??$3)、$??$E(-2，$??$0)$?
∴?$\begin{cases}{2k+b=3}\\{-2k+b=0}\end{cases}，$?解得?$\begin{cases}{k=\dfrac 34}\\{b=\dfrac 32}\end{cases}$?
∴直線?$CE$?的表達(dá)式為?$y=\frac {3}{4} x+ \frac {3}{2}$?
設(shè)點(diǎn)?$Q(t，$??$- \frac {3}{8}t^2+ \frac {3}{4}t+3 )$?
則點(diǎn)?$Q $?向左平移?$2$?個(gè)單位長(zhǎng)度，向上平移?$3$?個(gè)單位長(zhǎng)度得到點(diǎn)?$P(t-2，$??$-\frac {3}{8}t^2+ \frac {3}{4}t+6 )$?
將?$P(t-2，$??$-\frac {3}{8}t^2+ \frac {3}{4}t+6 )$?代入?$y=\frac {3}{4} x+ \frac {3}{2}$?
解得?$t_{1}=-4，$??$t_{2}=4($?舍去)
∴點(diǎn)?$Q $?的坐標(biāo)為?$(-4，$??$-6) $?
?$(3)$?將?$E(-2，$??$0)$?代入?$y=ax^2-2ax+c，$?得?$c=-8a$?
∴?$y=ax^2-2ax-8a=a(x-1)^2-9a$?
∴頂點(diǎn)坐標(biāo)為?$(1，$??$-9a)$?
① 當(dāng)拋物線頂點(diǎn)在正方形內(nèi)部時(shí)，與正方形有兩個(gè)交點(diǎn)
∴?$0< -9a< 3，$?解得?$-\frac {1}{3} < a< 0$?
②當(dāng)拋物線與直線?$BC$?的交點(diǎn)在點(diǎn)?$C$?上方，且與直線?$AD$?的交點(diǎn)在點(diǎn)?$D$?下方時(shí)
與正方形有兩個(gè)交點(diǎn)，即?$\begin{cases}{a+2a-8a< 3}\\{a×2^2-2a×2-8a> 3}\end{cases}$?
解得?$- \frac {3}{5} < a< - \frac {3}{8}$?
綜上所述，?$a$?的取值范圍為?$- \frac {1}{3} < a< 0$?或?$- \frac {3}{5} < a< - \frac {3}{8}$?
