$\left(1\right)\because \angle AOB=90^{\circ}$，$\angle ABO=30^{\circ}$，$AB=2$，
$\therefore OA=1$，$OB=\sqrt{3}$，
$\therefore A\left(0,1\right)$，$B(-\sqrt{3}$，$0)$，
$\because OB=OC$，
$\therefore OC=\sqrt{3}$，
$\therefore C(\sqrt{3}$，$0)$.
$(2)$过点$D$作$DM\bot y$轴于点$M$，过点$D$作$DN\bot x$轴于点$N$，

由题意，$y$轴是线段$BC$的垂直平分线，
$\therefore AB=AC$，
$\therefore \angle ABO=\angle ACO=30^{\circ}$，
$\because \angle DAC=90^{\circ}$，$x$轴$\bot y$轴，
$\therefore \angle DAM=\angle ACO=30^{\circ}$，
又$AD=AC$，$\angle AMD=\angle CAO$，
$\therefore \triangle AMD$≌$\triangle COA\left(AAS\right)$，
$\therefore DM=AO$，$AM=CO$，
$\because AO=1$，$CO=\sqrt{3}$，
$\therefore DM=ON=1$，$AM=\sqrt{3}$，
$\therefore D(1$，$\sqrt{3}+1)$，
$\therefore DN=\sqrt{3}+1$，
又$BN=OB+ON=\sqrt{3}+1$，
$\therefore DN=BN$，
$\therefore \triangle BND$是等腰直角三角形，
$\therefore \angle DBN=45^{\circ}$，
$\therefore \triangle GBO$是等腰直角三角形，
$\therefore BG=\sqrt{2}OB=\sqrt{2}×\sqrt{3}=\sqrt{6}$；
$(3)$由（2）可知：$\angle DBN=45^{\circ}$，$\angle DCB=30^{\circ}+45^{\circ}=75^{\circ}$，
$\therefore \angle BDC=180^{\circ}-45^{\circ}-75^{\circ}=60^{\circ}$，
$\because \angle BEC=\angle BDC$，
$\therefore \angle BEC=60^{\circ}$，
延长$EB$至$F$，使$BF=CE$，连接$AF$，

$\because \angle ABC=\angle ACB=30^{\circ}$，
$\therefore \angle BAC=120^{\circ}$，
$\therefore \angle ACE+\angle ABE=180^{\circ}$，
$\because \angle ABF+\angle ABE=180^{\circ}$，
$\therefore \angle ABF=\angle ACE$，
又$\because AB=AC$，$BF=CE$，
$\therefore \triangle ABF$≌$\triangle ACE\left(SAS\right)$，
$\therefore AF=AE$，$\angle BAF=\angle CAE$，
$\therefore \angle FAE=\angle BAC=120^{\circ}$，
$\therefore FE=\sqrt{3}AE$，
$\therefore BE+CE=BE+BF=FE=\sqrt{3}AE$，
即$BE+CE=\sqrt{3}AE$.

$\left(1\right)\because \angle AOB=90^{\circ}$，$\angle ABO=30^{\circ}$，$AB=2$，
$\therefore OA=1$，$OB=\sqrt{3}$，
$\therefore A\left(0,1\right)$，$B(-\sqrt{3}$，$0)$，
$\because OB=OC$，
$\therefore OC=\sqrt{3}$，
$\therefore C(\sqrt{3}$，$0)$.
$(2)$过点$D$作$DM\bot y$轴于点$M$，过点$D$作$DN\bot x$轴于点$N$，

由题意，$y$轴是线段$BC$的垂直平分线，
$\therefore AB=AC$，
$\therefore \angle ABO=\angle ACO=30^{\circ}$，
$\because \angle DAC=90^{\circ}$，$x$轴$\bot y$轴，
$\therefore \angle DAM=\angle ACO=30^{\circ}$，
又$AD=AC$，$\angle AMD=\angle CAO$，
$\therefore \triangle AMD$≌$\triangle COA\left(AAS\right)$，
$\therefore DM=AO$，$AM=CO$，
$\because AO=1$，$CO=\sqrt{3}$，
$\therefore DM=ON=1$，$AM=\sqrt{3}$，
$\therefore D(1$，$\sqrt{3}+1)$，
$\therefore DN=\sqrt{3}+1$，
又$BN=OB+ON=\sqrt{3}+1$，
$\therefore DN=BN$，
$\therefore \triangle BND$是等腰直角三角形，
$\therefore \angle DBN=45^{\circ}$，
$\therefore \triangle GBO$是等腰直角三角形，
$\therefore BG=\sqrt{2}OB=\sqrt{2}×\sqrt{3}=\sqrt{6}$；
$(3)$由（2）可知：$\angle DBN=45^{\circ}$，$\angle DCB=30^{\circ}+45^{\circ}=75^{\circ}$，
$\therefore \angle BDC=180^{\circ}-45^{\circ}-75^{\circ}=60^{\circ}$，
$\because \angle BEC=\angle BDC$，
$\therefore \angle BEC=60^{\circ}$，
延长$EB$至$F$，使$BF=CE$，连接$AF$，

$\because \angle ABC=\angle ACB=30^{\circ}$，
$\therefore \angle BAC=120^{\circ}$，
$\therefore \angle ACE+\angle ABE=180^{\circ}$，
$\because \angle ABF+\angle ABE=180^{\circ}$，
$\therefore \angle ABF=\angle ACE$，
又$\because AB=AC$，$BF=CE$，
$\therefore \triangle ABF$≌$\triangle ACE\left(SAS\right)$，
$\therefore AF=AE$，$\angle BAF=\angle CAE$，
$\therefore \angle FAE=\angle BAC=120^{\circ}$，
$\therefore FE=\sqrt{3}AE$，
$\therefore BE+CE=BE+BF=FE=\sqrt{3}AE$，
即$BE+CE=\sqrt{3}AE$.