$2$ can be done more easily in the manner of $1$:
There are $4$ pairs. We must include the pair $(0,1)$ in our relations. The remaining $3$ may or may not be in our relations as we choose. So there are $2^3$ such relations that most include $(0,1)$ but have no other requirement.
$3$ is a matter of looking at the eight relations and seeing which are reflexive, symmetric, antisymmetric and/or transitive.
You write "but in this case I don't see how I should apply the definition to the pairs above". I'm not sure I understand your confusion. You have $8$ relations and suddenly you are asking how to apply the definition of refl, symm, etc to the pairs. Why do you think we are apply them to pairs? I don't see why you made that assumpition.
Take the relations 1 at a time:
1) $\{(0,1)\}$.
This is not reflexive as it does not include both $(0,0)$ nor $(1,1)$.
This is not symmetric as for every $(a,b)$ contained (only $(0,1)$ is contained) then $(b,a)$ is not contained. ($(1,0)$ is not contain.)
It is antisymmetric as for any $(a,b); a\ne b$ contained (That's only $(0,1)$), then $(b,a)$ is not contained ($(1,0)$ is not contained).
And it is transitive as for any $(a,b)$ and $(b,c)$ contained (there are none; we have $(0,1)$ but we do not have $(1,x)$ at all), then $(a,c)$ is contained. This is vacuously true as there are no $(y,0)$ or $(1,x)$.
2) $ \{(0,1),(1,0)\}$
Not reflexive as it does not contain either $(0,0)$ nor $(1,1)$.
Is symmetric as for ever $(a,b)$ then $(b,a)$ is contained.
Not antisymetric as both $(1,0)$ and $(0,1)$ is contained.
Not transitive $(0,1)$ is contained and $(1,0)$ is contained but $(0,0)$ is not.
.....
And so on.
Six more to go.
====
Alternatively it will probably by easier to list the relations that have properties by the properties.
Reflexive.
$(a,a)$ is contained for all $a\in \{0,1\}$ so we must pick the relations that have both $(0,0)$ and $(1,1)$.
They are $\{(0,0),(1,1),(0,1)\}$ and $\{(0,0),(1,1),(0,1),(1,0)\}$
Symmetric.
To be symmetric if it has $(a,b)$ it must have $(b,a)$. Now all $(a,a)$ are symmetric to itself so if it has an $(a,a)$ we don't have to check for its converse $(a,a)$. But if it has a $(a,b)a\ne b$ we must check for $(b,a)$. And $(0,1)$ and $(1,0)$ are the only unequal pairs, a relation will be symmetric if and only if it has both $(0,1)$ and $(1,0)$ or it has neither. As we are only checking the relations with $(0,1)$ the we must find those that have both.
$\{(0,1),(1,0)\}$ and $\{(0,1),(1,0),(0,0)\}$ and $\{(0,1),(1,0),(1,0)\}$ and $\{(0,1),(1,0),(0,0),(1,1)\}$
Antisymmetric. If it has $(a,b) a\ne b$ it can not have $(b,a)$. As we are only considering relations with $(0,1)$ we must exclude $(1,0)$. As $0, 1$ are the only unequal elements that's all we have to check.
$\{(0,1)\}, \{(0,1),(0,0)\}, \{(0,1),(1,1)\}, \{(0,1),(0,0),(1,1)\}$.
Transitive. We have $(0,1)$. if we have any $(1,x)$ we must also have $(0,x)$ And if we have $(x,0)$ we must also have $(x,1)$. But we needn't have either of those.
So $\{(0,1)\}$. And $\{(0,1),(1,0),(0,0),(1,1)\}$. And $\{(0,1),(1,1)\}$ and $\{(0,1),(0,0)\}$.
The formula is:
$$t_{n,3} =2{n \choose 2} + 37{n \choose 3} + 116 {n \choose 4} + 180 {n \choose 5} + 120 {n \choose 6}$$
The general idea is to define $t_{n,3} = \sum_{k}d_{3,k}{n \choose k}$ where $d_{3,k}$ is the number of possible transitive relations in a set of size $k$ with exactly $3$ ordered pairs, that involve every element (ie. every element must appear in at least one pair). Then we generalize for a set of more than $k$ elements by doing a binomial of the subset of elements that do appear in pairs over the rest.
We have $3$ relations, involving no more than $6$ distinct elements, and the magic $d_{3,k}$ constants can be manually computed using a program. (tough it would be more elegant to find a solution that works without the aid of a computer).
Edit: here is the code I used to generate $d_{3,k}$ :
TransitiveMerge[{a_, b_}, {b_, c_}] := {{a, c}}
TransitiveMerge[{b_, c_}, {a_, b_}] := {{a, c}}
TransitiveMerge[x_, y_] := {}
TransitiveClosure[a_] :=
DeleteDuplicates[
Sort[Join[a,
Catenate[Map[TransitiveMerge[#[[1]], #[[2]]] &, Tuples[{a, a}]]]]]]
AllGraphs[n_, k_] := Subsets[Tuples[{Range[1, n], Range[1, n]}], {k}]
d[n_, k_] :=
d[n, k] =
Length[Select[
AllGraphs[n,
k], (TransitiveClosure[#] === #) &&
Length[DeleteDuplicates[Catenate[#]]] === n &]]
Comb[n_, k_] := Binomial[n, k] /; (NumberQ[n] && NumberQ[k])
t[n_, k_] := Apply[Plus, Map[Comb[n, #]* d[#, k] &, Range[1, 2*k]]]
t[n, 3]
t[n, 2]
Best Answer
You have:
4 different pairs with distinct numbers (i,j),(j,i),(k,l),(l,k) (out of the total of ${7\choose 2}=21$ pairs): ${21\choose 2}=210$
2 pairs with distinct elements and 2 pairs with same element (i,j),(j,i),(k,k),(l,l): ${7\choose 2}{7\choose 2}=441$
4 pairs with the same element (i,i),(j,j),(k,k),(l,l): ${7\choose 4}=35$
The total is 686
EDIT: At OP's request, here is a more systematic way of looking at the problem:
An element of your relation can come from one of the following two sources:
One can get $4$ elements in the relation in one of the following three ways:
So the total is ${21\choose2}{7\choose0}+{21\choose1}{7\choose2}+{21\choose0}{7\choose4}=686$
The general formula for $n$ elements and $k$ ordered pairs is:
$$\sum_{i=0}^{\lfloor k/2\rfloor}{m\choose i}{n\choose k-2i}\text{, where }m={n\choose 2}$$