For a relation $R$ to be symmetric, we have to have for all elements in $R$ that if $(x,y) \in R$, then also $(y,x) \in R$. You have found some elements in $R_1$ such that both $(x,y) \in R_1$ and $(y,x) \in R_1$, but for example $(2,4) \in R_1$ but $(4,2) \notin R_1$, hence it it not symmetric because it doesn't satisfy the criterion for every element.
There is also an element in $R_6$ that makes it non-symmetric, can you find it?
As for $R_5$, since every element is of the form $(x,x)$, it also (obviously) holds that $(x,x) \in R_6$, so it is symmetric.
The best and the most reliable order to satisfy properties of equivalence relation is in the given order
=> Reflexive Closure-->Symmetric Closure-->Transitivity closure
The reason for this assertion is that like for instance if you are following the order
=> Transitivity closure-->Reflexive Closure-->Symmetric Closure
(just like has been asked) ,you may just end up with elements that you added for symmetric closure not being accounted for transitivity as has been shown in the example given in question which has been cited here for reference
R={(2,1),(2,3)} .
Transitive closure: {(2,1),(2,3)}.
Reflexive closure: {(1,1),(2,1),(2,2),(2,3),(3,3)}.
Symmetric closure: {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}.
Since the set is missing (1,3) and (3,1) to be transitive, it is not an equivalence relation.
But if you follow the order of satisfying Reflexive Closure first,then Symmetric Closure and at last Transitivity closure,then the equivalence property is satisfied as shown.
R={(2,1),(2,3)} .
Reflexive closure: {(1,1),(2,1),(2,2),(2,3),(3,3)}.
Symmetric closure: {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}
Transitive closure:{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,2),(3,1),(3,3)}
Best Answer
$1$. The function you gave is symmetric, transitive and reflexive.
For a function that is symmetric, reflexive but not transitive take $\{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}$
$2$. the function you gave is not transitive since we have $(1,2)$ and $(2,3)$ but not $(1,3)$.
For an example that works take $(1,1),(2,2),(3,3),(1,2),(2,3),(1,3),(1,3)$. So just add $(1,3)$ to make it transitive and it works.
$3$. your example is good. You can also take $\emptyset$.