$Q1$
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
It looks like you misread slightly: partial orders and equivalence relations are both reflexive and transitive, but only equivalence relations are symmetric, while partial orders are antisymmetric. A relation $R$ is symmetric if $aRb$ implies $bRa$, while a relation $R$ is antisymmetric if $aRb$ and $bRA$ implies $a=b$.
For example, the relation "has the same parity as" is symmetric (and in fact an equivalence relation) on the integers, since if $a$ has the same parity as $b$ then $b$ has the same parity as $a$. On the other hand, it is not antisymmetric, because $2$ has the same parity as $4$ and $4$ has the same parity as $2$, even though $2\neq 4$.
The relation $\leq$ on integers, in contrast, is antisymmetric (and in fact a partial order), since if $a\leq b$ and $b\leq a$ then $a=b$. But it is not symmetric, since (for instance) $2\leq 4$ but $4\not\leq 2$.