$Q1$
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 is a partial order. We just need to check the three conditions:
$\underline{\text{Reflexivity}}$: we have $(a,a)$, $(b,b)$ and $(c,c)$, so it is reflexive.
$\underline{\text{Anti-symmetricity}}$: note that we don't have $(x,y)$ and $(y,x)$ for all $x,y \in \{a,b,c\}$, so this condition is trivially satisfied, so it is anti-symmetric.
$\underline{\text{Transitivity}}$: it is easy to check that this condition is satisfied, too. Let's do this carefully: $$(a,a) \ \text{and} \ (a,b) \rightarrow (a,b)$$ $$(a,a) \ \text{and} \ (a,c) \rightarrow (a,c)$$ $$(a,b) \ \text{and} \ (b,b) \rightarrow (b,b)$$ $$(a,c) \ \text{and} \ (c,c) \rightarrow (c,c)$$ This is enough to show that the relation is transitive. Why? All other possible combinations to check don't appear in our relation; for example $(a,b)$ and $(b,c)$ would need to imply $(a,c)$, but $(b,c)$ is not in our relation, so this condition is trivially satisfied.