Relations in set with 2 elements $$A = \{\{\sim\}, *\}$$
find 4 reflexive relations:
$$R_{1} = \{(\{\sim\}, \{\sim\})\}$$
$$R_{2} = \{(*, *)\}$$
$$R_{3} = \{(\{\sim\}, \{\sim\}), (*, *)\}$$
$$R_{4} = \{(\{\sim\}, \{\sim\}), (*, *), (\{\sim\}, *)\}$$
find 2 symmetric relations:
$$R_{1} = \{(\{\sim\}, *), (*, \{\sim\})\}$$
$$R_{2} = \{(*, *)\}$$
find 1 transitive relation:
$$R = \{(\{\sim\}, *), (*, \{\sim\}), (\{\sim\}, \{\sim\})\}$$
I'm not sure about that because of 2 elements in set $A$.
 A: Big picture.  A relationship on $A$ is any subset set of $P(A\times A)$.
Although this isn't asked for $|A\times A| = 2\times 2 = 4$ and $|P(A\times A)| = 2^4 =16$ so there are exactly $16$ possible relations.
1) Reflexive:
That means for every $a \in A$ then $(a,a) \in R$.
So $(\{\sim\},\{\sim\}), (*,*) \in R$ all other ordered pairs ($(\{\sim\},*)$ and $(*,\{\sim\})$)  may or may not be.
There are $4$ options as to whether $(\{\sim\},*)$ are  $(*,\{\sim\})$ in $R$ so there are $4$ reflexive relations.
They are 
$\{(\{\sim\},\{\sim\}), (*,*)\}$
$\{(\{\sim\},\{\sim\}), (*,*),(\{\sim\},*)\}$
$\{(\{\sim\},\{\sim\}), (*,*),(*,\{\sim\})\}$
$\{(\{\sim\},\{\sim\}), (*,*),(\{\sim\},*)(*,\{\sim\})\}$
2) Symmetric.  
Means if $(a,b)\in R$ then $(b,a)\in R$.  So if $(*,\{\sim\})\in R$ then $(\{\sim\},*)$ must be also.  But neither have to be.  (Obviously if any $(a,a)\in R$ then.....$(a,a)\in R$ and we needn't worry about twin pairs.)
So of the pairs.  $(*,*)$, $(\{\sim\},\{\sim\})$ and the "glued" pair of pairs $(*,\{\sim\})|(\{\sim\},*)$ each may or may not be in $R$ so there are $2^3 =8$ possible symmetric pairs.
They are:
$\emptyset$, $\{(*,*)\}$, $\{(\{\sim\},\{\sim\})\}$,  $\{(*,*),(\{\sim\},\{\sim\})\}$ and
$\{(*,\{\sim\}),(\{\sim\},*)\},\{(*,\{\sim\}),(\{\sim\},*),(*,*)\}, \{(*,\{\sim\}),(\{\sim\},*),(\{\sim\},\{\sim\})\} ,\{(*,\{\sim\}),(\{\sim\},*), (*,*),(\{\sim\},\{\sim\})\}$
3) Transitive....
Okay... a different tack.
If we have no $(a,b);a\ne b$ in $R$ then $r$ must be transitive as the only we we can have $(a,b)$ and $(b,c)\in R$ is if $a=b$ and $b=c$ and if $(a,a)\in R$ then .... $(a,a)\in R$.
So of those type we may or may not have $(*,*)$ or $(\{\sim\},\{\sim\})$ in $R$ so there are $4$ if these vacuous like transitive relation.
$\emptyset$,$\{(*,*)\}$, $\{(\{\sim\},\{\sim\})\}$, and $\{(*,*),(\{\sim\},\{\sim\})\}$
If we have $(a,b)$ but not $(b,a)$ then well that's transitive too. If $(a,a)$ is or is not in $R$ then $(a,b)$ is in $R$ and if $(b,b)$ is or is not in $R$ then $(a,b)$ is in $R$.
So there are $2$ choices for $(a,b);a\ne b$ and $4$ options if the twin pairs are include or not so there are $8$ of these type function.
$\{(*,\{\sim\})\}$, $\{(*,*),(*,\{\sim\})\}$, $\{(\{\sim\},\{\sim\}),(*,\{\sim\})\}$, and $\{(*,*),(\{\sim\},\{\sim\}),(*,\{\sim\})\}$
And 
$\{(\{\sim\},*)\}$,$\{(*,*),(\{\sim\},*)\}$, $\{(\{\sim\},\{\sim\}),(\{\sim\},*)\}$, and $\{(*,*),(\{\sim\},\{\sim\}),(\{\sim\},*)\}$
ANd if we have both $(a,b)$ and $(b,a)$ then we must have $(a,a)$ and $(b,b)$ and there is one that will do that
$\{(*,*),(*,\{\sim\}), (\{\sim\},*),(\{\sim\},\{\sim\})$.
There are $13$ transitive relations so only $3$ that aren't!
They are the ones containing both $(a,b)$ and $(b,a)$ but not both $(a,a)$ and $(b,b)$.
...
The question didn't ask but of the $4$ reflexive, $8$ symmetric and $13$ transitive the only ones that are all three are 
$\{(\{\sim\},\{\sim\}), (*,*)\}$ and $\{(\{\sim\},\{\sim\}), (*,*),(\{\sim\},*)(*,\{\sim\})\}$
Those are the two equivalence relationships.
The first has two equivalence classes:  $C_1 = \{*\}$ and $C_2= \{\{\sim\}\}$.  i.e elements are only related to themselves.
The other has one equivalence class  $C= \{*,\{\sim\}\} = A$. i.e. Everything is related to everything.
A: A reflexive relation over $\{\{\sim\},\ast\}$ must include at least both $\langle\{\sim\},\{\sim\}\rangle$ and $\langle\ast,\ast\rangle$ as elements.  
Try again.  Only $R_3$ and $R_4$ are reflexive.

You have two symmetric relations.

In a transitive relation, if $\langle\{\sim\},\ast\rangle$ and $\langle\ast,\{\sim\}\rangle$ are included, so to must $\langle\{\sim\},\{\sim\}\rangle$. 
In a transitive relation, if $\langle\ast,\{\sim\}\rangle$ and $\langle\{\sim\},\ast\rangle$ are included ...
