Being a reflexive relation is an extrinsic property. It connects the relation with an external set, while being symmetric and transitive are intrinsic properties that depend only on the ordered pairs in the relation.
$R$ is reflexive on $A$ if for every $x\in A$ the pair $\langle x,x\rangle$ is in $R$. So if $R=\varnothing$ and $A\neq\varnothing$ it is not reflexive on $A$.
The same would hold whether $\{\langle 1,1\rangle\}$ is reflexive on $\{1\}$? Yes. It is, but is it reflexive on $\Bbb N$? No. The pair $\langle 2,2\rangle$ is not there, so not every $n\in\Bbb N$ satisfies the property required.
$Q1$
Can a relation be both partial order and equivalence?
Yes, for example, the equality relation.
Is $R_1$ Transitive?
No. It has $(1,0)$ and $(0,7)$ but not $(1,7)$. As this example show, if you add an ordered pair to a transitive relation it can become non-transitive.
A relation on set $A$ that is both reflexive and transitive but neither an equivalence relation nor a partial order (meaning it is neither symmetric nor antisymmetric) is:
$$R_3 = \left\{(0,0),\, (7,7),\, (1,1),\, (0,7),\, (7,1),\, (0,1),\, (1,7) \right\}$$
Reflexive? Yes, because it has $(0,0),\, (7,7),\, (1,1)$.
Transitive? Yes. We go through the relevant cases:
$$(0,7) \mbox{ and } (7,1) \Rightarrow (0,1) \qquad\checkmark$$
$$(7,1) \mbox{ and } (1,7) \Rightarrow (7,7) \qquad\checkmark$$
$$(0,1) \mbox{ and } (1,7) \Rightarrow (0,7) \qquad\checkmark$$
$$(1,7) \mbox{ and } (7,1) \Rightarrow (1,1) \qquad\checkmark$$
Symmetric? No, because we have $(0,1)$ but not $(1,0)$
Antisymmetric? No, because we have $(1,7)$ and $(7,1)$.
$Q2$
Your relation, $R_2$, is correct but your explanations for symmetric and antisymmetric are the wrong way around.
$R_2$ is not antisymmetric because there is as two-way street between distinct vertices, namely, $0$ and $7$.
$R_2$ is symmetric because there is no one-way street between distinct vertices.
Also, $R_2$ is not transitive because it has $(0,7)$ and $(7,0)$ but not $(0,0)$.
Best Answer
HINT: $S\times S$ is the set of all ordered pairs of sequences of real numbers; it is indeed infinite, since $S$ is. The relation $R$ has all but one of the properties of reflexivity, transitivity, antisymmetry, and symmetry, but this has nothing to do with pairing real numbers: members of $R$ are pairs of sequences of real numbers, not pairs of real numbers.
If $a=\langle a_n:n\in\Bbb N\rangle$ and $b=\langle b_n:n\in\Bbb N\rangle$ are two sequences of real numbers, the pair $\langle a,b\rangle$ is in $R$ if and only if $a_3=b_3$. For example, if $a$ is the constant sequence defined by $a_n=3$ for all $n\in\Bbb N$, and $b$ is the sequence defined by $b_n=n$ for all $n\in\Bbb N$, then $\langle a,b\rangle=R$, because $a_3=b_3=3$.
Reflexivity: If $a=\langle a_n:n\in\Bbb N\rangle$ is a sequence of real numbers, is it true that $a_3=a_3$? If this is always the case, then $\langle a,a\rangle\in R$ for each $a\in S$, and $R$ is therefore reflexive. If there is even one sequence in $S$ for which it’s not true, then $R$ is not reflexive.
Symmetry: If $a=\langle a_n:n\in\Bbb N\rangle$ and $b=\langle b_n:n\in\Bbb N\rangle$ are sequences of real numbers, and $\langle a,b\rangle\in R$, is it always true that $\langle b,a\rangle\in R$? In other words, if $a_3=b_3$, is it always true that $b_3=a_3$?
Transitivity: If $a=\langle a_n:n\in\Bbb N\rangle$, $b=\langle b_n:n\in\Bbb N\rangle$, and $c=\langle c_n:n\in\Bbb N\rangle$ are sequences of real numbers, $\langle a,b\rangle\in R$, and $\langle b,c\rangle\in R$, is it always true that $\langle a,c\rangle\in R$? Use the definition of $R$ as I did in the first two parts to translate this question into a more basic one.
Antisymmetry: If $a=\langle a_n:n\in\Bbb N\rangle$ and $b=\langle b_n:n\in\Bbb N\rangle$ are sequences of real numbers, $\langle a,b\rangle\in R$, and $\langle b,a\rangle\in R$, is it always true that $a=b$? Again, use the definition of $R$ as I did in the first two parts to translate this question into a more basic one.