$\left(1\right)\because$抛物线$y=ax^{2}+bx+c$经过点$A\left(-1,2\right)$、$B\left(2,2\right)$、$C\left(3,-2\right)$，
$\therefore \left\{\begin{array}{l}{a-b+c=2}\\{4a+2b+c=2}\\{9a+3b+c=-2}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=-1}\\{b=1}\\{c=4}\end{array}\right.$，
$\therefore$此抛物线的解析式为$y=-x^{2}+x+4$；
$(2)$设直线$AC$的解析式为$y=kx+b$，
$\therefore \left\{\begin{array}{l}{-k+b=2}\\{3k+b=-2}\end{array}\right.$，
解得$\left\{\begin{array}{l}{k=-1}\\{b=1}\end{array}\right.$，
$\therefore$直线$AC$的解析式为$y=-x+1$，
设$P(m$，$-m^{2}+m+4)$，过点$P$作$PH\bot x$轴，交$AC$于$H$，如图$1$，

则$H\left(m,-m+1\right)$，
$\therefore PH=-m^{2}+m+4-\left(-m+1\right)=-m^{2}+2m+3$，
$\because S_{\triangle ACP}=\frac{1}{2}\times PH\times (x_{C}-x_{A})=\frac{1}{2}(-m^{2}+2m+3)\times 4=-2m^{2}+4m+6=6$，
解得：$m_{1}=0$，$m_{2}=2$，
$\therefore P$点的坐标为$\left(0,4\right)$或$\left(2,2\right)$；
$(3)\because$直线$y=-\frac{1}{2}x+t$分别交$x$轴、$y$轴于点$D$、$E$.
$\therefore D\left(2t,0\right)$，$E\left(0,t\right)$，且$t \lt 0$，
$\therefore OD=-2t$，$OE=-t$，
$\because \angle EDF=\angle DOE=\angle FOD=90^{\circ}$，
$\therefore \angle FDO+\angle EDO=\angle EDO+\angle DEO=90^{\circ}$，
$\therefore \angle FDO=\angle DEO$，
$\therefore \triangle DFO$∽$\triangle EDO$，
$\therefore \frac{OF}{OD}=\frac{OD}{OE}$，即$\frac{OF}{-2t}=\frac{-2t}{-t}$，
$\therefore OF=-4t$，
$\therefore F\left(0,-4t\right)$，
$\therefore$直线$DF$的解析式为$y=2x-4t$，
联立得$\left\{\begin{array}{l}{y=2x-4t}\\{y=-{x}^{2}+x+4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{{x}_{1}=\frac{-1-\sqrt{16t+17}}{2}}\\{{y}_{1}=-1-\sqrt{16t+17}-4t}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=\frac{-1+\sqrt{16t+17}}{2}}\\{{y}_{2}=-1+\sqrt{16t+17}-4t}\end{array}\right.$，
$\therefore Q_{1}(\frac{-1-\sqrt{16t+17}}{2}$，$-1-\sqrt{16t+17}-4t)$，$Q_{2}(\frac{-1+\sqrt{16t+17}}{2}$，$-1+\sqrt{16t+17}-4t)$，
$\therefore Q_{1}D=\sqrt{(\frac{-1-\sqrt{16t+17}}{2}-2t)^{2}+(-1-\sqrt{16t+17}-4t)^{2}}=\frac{\sqrt{5}}{2}(-1-4t-\sqrt{16t+17})$，
同理可得：$Q_{2}D=\frac{\sqrt{5}}{2}(-1-4t+\sqrt{16t+17})$，
在$Rt\triangle DEO$中，$DE=\sqrt{O{D}^{2}+O{E}^{2}}=\sqrt{(-2t)^{2}+(-t)^{2}}=-\sqrt{5}t$，
当$\triangle EDQ_{1}$∽$\triangle DOE$时，如图$2$，则$\frac{D{Q}_{1}}{OE}=\frac{DE}{OD}$，

$\therefore DQ_{1}\cdot OD=DE\cdot OE$，
即$\frac{\sqrt{5}}{2}(-1-4t-\sqrt{16t+17})\times \left(-2t\right)=-\sqrt{5}t\cdot \left(-t\right)$，
解得：$t=2$或$t=-\frac{8}{9}$，
经检验，$t=2$或$t=-\frac{8}{9}$均是原方程的解，
$\because t \lt 0$，
$\therefore t=-\frac{8}{9}$；
当$\triangle Q_{2}DE$∽$\triangle DOE$时，则$\frac{D{Q}_{2}}{DE}=\frac{OD}{OE}$，
$\therefore DQ_{2}\cdot OE=OD\cdot DE$，
即$\frac{\sqrt{5}}{2}(-1-4t+\sqrt{16t+17})\times \left(-t\right)=-2t\cdot (-\sqrt{5}t)$，
解得：$t=-1$，
经检验，$t=-1$是原方程的解，
$\therefore t=-1$；
综上所述，抛物线上存在点$Q$，使得以$D$为直角顶点的$\triangle QDE$与$\triangle DOE$相似，$t$的值为$-\frac{8}{9}$或$-1$.

$\left(1\right)\because$抛物线$y=ax^{2}+bx+c$经过点$A\left(-1,2\right)$、$B\left(2,2\right)$、$C\left(3,-2\right)$，
$\therefore \left\{\begin{array}{l}{a-b+c=2}\\{4a+2b+c=2}\\{9a+3b+c=-2}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{a=-1}\\{b=1}\\{c=4}\end{array}\right.$，
$\therefore$此抛物线的解析式为$y=-x^{2}+x+4$；
$(2)$设直线$AC$的解析式为$y=kx+b$，
$\therefore \left\{\begin{array}{l}{-k+b=2}\\{3k+b=-2}\end{array}\right.$，
解得$\left\{\begin{array}{l}{k=-1}\\{b=1}\end{array}\right.$，
$\therefore$直线$AC$的解析式为$y=-x+1$，
设$P(m$，$-m^{2}+m+4)$，过点$P$作$PH\bot x$轴，交$AC$于$H$，如图$1$，

则$H\left(m,-m+1\right)$，
$\therefore PH=-m^{2}+m+4-\left(-m+1\right)=-m^{2}+2m+3$，
$\because S_{\triangle ACP}=\frac{1}{2}\times PH\times (x_{C}-x_{A})=\frac{1}{2}(-m^{2}+2m+3)\times 4=-2m^{2}+4m+6=6$，
解得：$m_{1}=0$，$m_{2}=2$，
$\therefore P$点的坐标为$\left(0,4\right)$或$\left(2,2\right)$；
$(3)\because$直线$y=-\frac{1}{2}x+t$分别交$x$轴、$y$轴于点$D$、$E$.
$\therefore D\left(2t,0\right)$，$E\left(0,t\right)$，且$t \lt 0$，
$\therefore OD=-2t$，$OE=-t$，
$\because \angle EDF=\angle DOE=\angle FOD=90^{\circ}$，
$\therefore \angle FDO+\angle EDO=\angle EDO+\angle DEO=90^{\circ}$，
$\therefore \angle FDO=\angle DEO$，
$\therefore \triangle DFO$∽$\triangle EDO$，
$\therefore \frac{OF}{OD}=\frac{OD}{OE}$，即$\frac{OF}{-2t}=\frac{-2t}{-t}$，
$\therefore OF=-4t$，
$\therefore F\left(0,-4t\right)$，
$\therefore$直线$DF$的解析式为$y=2x-4t$，
联立得$\left\{\begin{array}{l}{y=2x-4t}\\{y=-{x}^{2}+x+4}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{{x}_{1}=\frac{-1-\sqrt{16t+17}}{2}}\\{{y}_{1}=-1-\sqrt{16t+17}-4t}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=\frac{-1+\sqrt{16t+17}}{2}}\\{{y}_{2}=-1+\sqrt{16t+17}-4t}\end{array}\right.$，
$\therefore Q_{1}(\frac{-1-\sqrt{16t+17}}{2}$，$-1-\sqrt{16t+17}-4t)$，$Q_{2}(\frac{-1+\sqrt{16t+17}}{2}$，$-1+\sqrt{16t+17}-4t)$，
$\therefore Q_{1}D=\sqrt{(\frac{-1-\sqrt{16t+17}}{2}-2t)^{2}+(-1-\sqrt{16t+17}-4t)^{2}}=\frac{\sqrt{5}}{2}(-1-4t-\sqrt{16t+17})$，
同理可得：$Q_{2}D=\frac{\sqrt{5}}{2}(-1-4t+\sqrt{16t+17})$，
在$Rt\triangle DEO$中，$DE=\sqrt{O{D}^{2}+O{E}^{2}}=\sqrt{(-2t)^{2}+(-t)^{2}}=-\sqrt{5}t$，
当$\triangle EDQ_{1}$∽$\triangle DOE$时，如图$2$，则$\frac{D{Q}_{1}}{OE}=\frac{DE}{OD}$，

$\therefore DQ_{1}\cdot OD=DE\cdot OE$，
即$\frac{\sqrt{5}}{2}(-1-4t-\sqrt{16t+17})\times \left(-2t\right)=-\sqrt{5}t\cdot \left(-t\right)$，
解得：$t=2$或$t=-\frac{8}{9}$，
经检验，$t=2$或$t=-\frac{8}{9}$均是原方程的解，
$\because t \lt 0$，
$\therefore t=-\frac{8}{9}$；
当$\triangle Q_{2}DE$∽$\triangle DOE$时，则$\frac{D{Q}_{2}}{DE}=\frac{OD}{OE}$，
$\therefore DQ_{2}\cdot OE=OD\cdot DE$，
即$\frac{\sqrt{5}}{2}(-1-4t+\sqrt{16t+17})\times \left(-t\right)=-2t\cdot (-\sqrt{5}t)$，
解得：$t=-1$，
经检验，$t=-1$是原方程的解，
$\therefore t=-1$；
综上所述，抛物线上存在点$Q$，使得以$D$为直角顶点的$\triangle QDE$与$\triangle DOE$相似，$t$的值为$-\frac{8}{9}$或$-1$.