$\left(1\right)\because$点$E$是线段$AC$，$CD$的垂直平分线的交点，
$\therefore EA=EC=ED$，
$\therefore \angle EAC=\angle ECA$，$\angle ECD=\angle EDC$，
$\because \angle ABC=90^{\circ}$，$\angle BAC=50^{\circ}$，
$\therefore \angle ACB=90^{\circ}-50^{\circ}=40^{\circ}$，
$\therefore \angle ACD=180^{\circ}-40^{\circ}=140^{\circ}$，
$\therefore \angle EAC+\angle ACD+\angle EDC=280^{\circ}$，
$\therefore \angle AED=360^{\circ}-280^{\circ}=80^{\circ}$.
$(2)$①$\because$点$E$是线段$AC$，$CD$的垂直平分线的交点，
$\therefore EA=EC=ED$，
$\therefore \angle EAC=\angle ECA$，$\angle ECD=\angle EDC$，
$\because \angle ABC=90^{\circ}$，$\angle BAC=60^{\circ}$，
$\therefore \angle ACB=90^{\circ}-60^{\circ}=30^{\circ}$，
$\therefore \angle ACD=180^{\circ}-30^{\circ}=150^{\circ}$，
$\therefore \angle EAC+\angle ACD+\angle EDC=300^{\circ}$，
$\therefore \angle AED=360^{\circ}-300^{\circ}=60^{\circ}$，
$\therefore \triangle ADE$是等边三角形；
②点$P$在$ED\’$的延长线上时，$PE-PD$的值最大，理由如下：
如图$3$中，作点$D$关于直线$CF$的对称点${D\’}$，连接$CD\’$，$DD\’$，$ED\’$.

当点$P$在$ED\’$的延长线上时，$PE-PD$的值最大，此时$PE-PD=ED\’$，
$\because \angle CFD+\angle CFE=180^{\circ}$，$\angle CFD=\angle CAE$，
$\therefore \angle CAE+\angle CFE=180^{\circ}$，
$\therefore \angle ACF+\angle AEF=180^{\circ}$，
$\because \angle AED=60^{\circ}$，
$\therefore \angle ACF=120^{\circ}$，
$\therefore \angle ACB=\angle FCD=30^{\circ}$，
$\therefore \angle DCF=\angle FCD\’=30^{\circ}$，
$\therefore \angle DCD\’=60^{\circ}$，
$\because CD=CD\’$，
$\therefore \triangle CDD\’$是等边三角形，
$\therefore DC=DD\’$，$\angle CDD\’=\angle ADE=60^{\circ}$，
$\therefore \angle ADC=\angle EDD\’$，
$\because DA=DE$，
$\therefore \triangle ADC$≌$\triangle EDD\’\left(SAS\right)$，
$\therefore AC=ED\’$，
$\because \angle B=90^{\circ}$，$\angle ACB=30^{\circ}$，
$\therefore AC=2AB$，
$\therefore PE-PD=2AB=2\times 1=2$.
$\therefore$点$P$在$ED\’$的延长线上时，$PE-PD$的值最大，最大值为$2$，

$\left(1\right)\because$点$E$是线段$AC$，$CD$的垂直平分线的交点，
$\therefore EA=EC=ED$，
$\therefore \angle EAC=\angle ECA$，$\angle ECD=\angle EDC$，
$\because \angle ABC=90^{\circ}$，$\angle BAC=50^{\circ}$，
$\therefore \angle ACB=90^{\circ}-50^{\circ}=40^{\circ}$，
$\therefore \angle ACD=180^{\circ}-40^{\circ}=140^{\circ}$，
$\therefore \angle EAC+\angle ACD+\angle EDC=280^{\circ}$，
$\therefore \angle AED=360^{\circ}-280^{\circ}=80^{\circ}$.
$(2)$①$\because$点$E$是线段$AC$，$CD$的垂直平分线的交点，
$\therefore EA=EC=ED$，
$\therefore \angle EAC=\angle ECA$，$\angle ECD=\angle EDC$，
$\because \angle ABC=90^{\circ}$，$\angle BAC=60^{\circ}$，
$\therefore \angle ACB=90^{\circ}-60^{\circ}=30^{\circ}$，
$\therefore \angle ACD=180^{\circ}-30^{\circ}=150^{\circ}$，
$\therefore \angle EAC+\angle ACD+\angle EDC=300^{\circ}$，
$\therefore \angle AED=360^{\circ}-300^{\circ}=60^{\circ}$，
$\therefore \triangle ADE$是等边三角形；
②点$P$在$ED\’$的延长线上时，$PE-PD$的值最大，理由如下：
如图$3$中，作点$D$关于直线$CF$的对称点${D\’}$，连接$CD\’$，$DD\’$，$ED\’$.

当点$P$在$ED\’$的延长线上时，$PE-PD$的值最大，此时$PE-PD=ED\’$，
$\because \angle CFD+\angle CFE=180^{\circ}$，$\angle CFD=\angle CAE$，
$\therefore \angle CAE+\angle CFE=180^{\circ}$，
$\therefore \angle ACF+\angle AEF=180^{\circ}$，
$\because \angle AED=60^{\circ}$，
$\therefore \angle ACF=120^{\circ}$，
$\therefore \angle ACB=\angle FCD=30^{\circ}$，
$\therefore \angle DCF=\angle FCD\’=30^{\circ}$，
$\therefore \angle DCD\’=60^{\circ}$，
$\because CD=CD\’$，
$\therefore \triangle CDD\’$是等边三角形，
$\therefore DC=DD\’$，$\angle CDD\’=\angle ADE=60^{\circ}$，
$\therefore \angle ADC=\angle EDD\’$，
$\because DA=DE$，
$\therefore \triangle ADC$≌$\triangle EDD\’\left(SAS\right)$，
$\therefore AC=ED\’$，
$\because \angle B=90^{\circ}$，$\angle ACB=30^{\circ}$，
$\therefore AC=2AB$，
$\therefore PE-PD=2AB=2\times 1=2$.
$\therefore$点$P$在$ED\’$的延长线上时，$PE-PD$的值最大，最大值为$2$，