(i) From the figure we know that $$\triangle EDC$$ is an equilateral triangle
so we get $$\angle EDC=\angle ECD={60}^{o}$$
We know that $$ABCD$$ is a square
so we get $$\angle CDA=\angle DCB={90}^{o}$$
consider $$\triangle EDA$$
we get
$$\angle EDA={60}^{o}+{90}^{o}$$
so we get
$$\angle EDA={150}^{o}....(1)$$
Consider $$\triangle ECB$$
we get
$$\angle ECB=\angle ECD+\angle DCB$$
by substituting the values in the above equation
$$\angle ECB={60}^{o}+{90}^{o}$$
we get
$$\angle ECB={150}^{o}$$
so we get $$\angle EDA=\angle ECB....(ii)$$
consider $$\triangle EDA$$ and $$\triangle ECB$$
$$ED=EC$$ (sides of an equilateral triangle)
$$\angle EDA=\angle ECB$$ (From $$(ii)$$)
$$DA=CB$$ (sides of square)
by SAS congruence criterion
$$\triangle EDA\cong \triangle ECB$$
$$AE=BE$$
(ii) consider $$\triangle EDA$$
we know that
$$ED=DA$$
from the figure we know that the base angles are equal
$$\angle DEA=\angle DAE$$
based on equation (i) we get $$\angle EDA={150}^{o}$$
by angle sum property
$$\angle EDA+\angle DAE+\angle DEA={180}^{o}$$
by substituting the values we get
$${150}^{o}+\angle DAE+\angle DEA={ 180 }^{ o }$$
we know that $$\angle DEA=\angle DAE$$
so we get
$${150}^{o}+\angle DAE+\angle DAE={ 180 }^{ o }$$
$$2\angle DEA={ 180 }^{ o }-{150}^{o}$$
$$\angle DAE={15}^{o}$$