$\left(1\right)\because |a-6|+\sqrt{b-9}=0$，
又$\because |a-6|\geqslant 0$，$\sqrt{b-9}≥0$，
$\therefore a-6=0$，$b-9=0$，
解得$a=6$，$b=9$，
$\therefore A\left(0,6\right)$，$B\left(9,0\right)$；
$(2)$如图$1$，分别过点$B$，$A$作$x$轴，$y$轴的垂线交于点$H$，过点$C$作$CG\bot AH$于$G$，

$\because A\left(0,6\right)$，$B\left(9,0\right)$，$C\left(-3,n\right)$，
$\therefore B H=6$，$CG=6-n$，$AG=3$，$AH=9$，$GH=9-\left(-3\right)=12$，
$\because S_{\triangle ABC}=S_{梯形BHGC}-S_{\triangle ACG}-S_{\triangle ABH}=45$，
$\therefore \frac{1}{2}(BH+CG)•GH-\frac{1}{2}CG•AG-\frac{1}{2}AH•BH=45$，
即$\frac{1}{2}×(6+6-n)×12-\frac{1}{2}×(6-n)×3-\frac{1}{2}×9×6=45$，
解得$n=-2$，
$\therefore C\left(-3,-2\right)$，
$\therefore$点$A\left(0,6\right)$向左移动$3$个单位长度，向下移动$8$个单位长度得到点$C\left(-3,-2\right)$，
$\because$点$M\left(m,4\right)$在线段$AB$上，其对应点为$E\left(0,t\right)$，
$\therefore m=3$，$t=-4$；
$(3)\angle COF+\angle OFB=4\angle P$，理由如下：
如图$2$，过点$O$作$ON$∥$AB,$交$FP$于点$N$，过点$P$作$PM$∥$AB$，交$y$轴于点$M$，

设$\angle OFP=\alpha$，$\angle PCD=\beta$，
$\because FP$平分$\angle BFO$，$\angle BCP=3\angle PCD$，
$\therefore \angle OFP=\angle BFP=\alpha$，$\angle BCP=3\angle PCD=3\beta$，
$\because ON$∥$AB$，
$\therefore \angle ONF=\angle BFP=\alpha$，
$\because ON$∥$AB,PM$∥$AB$，
$\therefore ON$∥$PM$，
$\therefore \angle MPF=\angle ONF=\alpha$，
由平移的性质可得$,AB$∥$CD$，
$\therefore PM$∥$CD$，
$\therefore \angle MPC=\angle PCD=\beta$，
$\therefore \angle CPF=\angle MPC+\angle MPF=\alpha +\beta$，
$\because AB$∥$CD$，
$\therefore \angle OBF=\angle OCD=\angle PCD+\angle BCP=4\beta$，
又$\because \angle OFB=\angle OFP+\angle BFP=2\alpha$，
$\therefore \angle COF=\angle OFB+\angle OBF=2\alpha +4\beta$，
$\therefore \angle COF+\angle OFB=4\alpha +4\beta$，
$\therefore \angle COF+\angle OFB=4\angle P$.

$\left(1\right)\because |a-6|+\sqrt{b-9}=0$，
又$\because |a-6|\geqslant 0$，$\sqrt{b-9}≥0$，
$\therefore a-6=0$，$b-9=0$，
解得$a=6$，$b=9$，
$\therefore A\left(0,6\right)$，$B\left(9,0\right)$；
$(2)$如图$1$，分别过点$B$，$A$作$x$轴，$y$轴的垂线交于点$H$，过点$C$作$CG\bot AH$于$G$，

$\because A\left(0,6\right)$，$B\left(9,0\right)$，$C\left(-3,n\right)$，
$\therefore B H=6$，$CG=6-n$，$AG=3$，$AH=9$，$GH=9-\left(-3\right)=12$，
$\because S_{\triangle ABC}=S_{梯形BHGC}-S_{\triangle ACG}-S_{\triangle ABH}=45$，
$\therefore \frac{1}{2}(BH+CG)•GH-\frac{1}{2}CG•AG-\frac{1}{2}AH•BH=45$，
即$\frac{1}{2}×(6+6-n)×12-\frac{1}{2}×(6-n)×3-\frac{1}{2}×9×6=45$，
解得$n=-2$，
$\therefore C\left(-3,-2\right)$，
$\therefore$点$A\left(0,6\right)$向左移动$3$个单位长度，向下移动$8$个单位长度得到点$C\left(-3,-2\right)$，
$\because$点$M\left(m,4\right)$在线段$AB$上，其对应点为$E\left(0,t\right)$，
$\therefore m=3$，$t=-4$；
$(3)\angle COF+\angle OFB=4\angle P$，理由如下：
如图$2$，过点$O$作$ON$∥$AB,$交$FP$于点$N$，过点$P$作$PM$∥$AB$，交$y$轴于点$M$，

设$\angle OFP=\alpha$，$\angle PCD=\beta$，
$\because FP$平分$\angle BFO$，$\angle BCP=3\angle PCD$，
$\therefore \angle OFP=\angle BFP=\alpha$，$\angle BCP=3\angle PCD=3\beta$，
$\because ON$∥$AB$，
$\therefore \angle ONF=\angle BFP=\alpha$，
$\because ON$∥$AB,PM$∥$AB$，
$\therefore ON$∥$PM$，
$\therefore \angle MPF=\angle ONF=\alpha$，
由平移的性质可得$,AB$∥$CD$，
$\therefore PM$∥$CD$，
$\therefore \angle MPC=\angle PCD=\beta$，
$\therefore \angle CPF=\angle MPC+\angle MPF=\alpha +\beta$，
$\because AB$∥$CD$，
$\therefore \angle OBF=\angle OCD=\angle PCD+\angle BCP=4\beta$，
又$\because \angle OFB=\angle OFP+\angle BFP=2\alpha$，
$\therefore \angle COF=\angle OFB+\angle OBF=2\alpha +4\beta$，
$\therefore \angle COF+\angle OFB=4\alpha +4\beta$，
$\therefore \angle COF+\angle OFB=4\angle P$.