$Q1$
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 $(0,0)$.
Your intuition is correct. To describe the equivalence classes explicitly, it often helps to find the equivalence classes of numbers that are easy to work with.
Let's find, for example, the equivalence classes $[0]$ and $[1]$.
$0+a$ is even for which integers $a$? All these will form $[0]$.
$1+b$ is even for which integers $b$? All these will form $[1]$.
Are there any other equivalence classes? (You can be sure you've found them all when your equivalence classes form a partition of $\Bbb Z$.)
Best Answer
What is meant by "modulo operator is an equivalence relation" is the following:
This definition states in a mathematically precise way that $x \equiv y \pmod n$ if $x$ and $y$ have the same remainder modulo $n$.
Can you now prove that this $\equiv \pmod n$ is an equivalence relation? It is a good exercise to familiarise yourself with the concept.
Edit: It just occurred to me that you may be subconsciously bracketing the expression $x \equiv y \pmod 7$ in an unintended way. What is meant is:
$$(x \equiv y) \pmod 7$$
as opposed to:
$$x = (y \mathrel\% 7)$$
where $\%$ is the remainder operation. The former will be an equivalence relation. The latter won't, for $(7,7) \notin R$. I hope that clears the air for you.
The first notation $x \equiv y \pmod 7$ can alternatively be read as:
$$(x \mathrel\% 7) = (y \mathrel\% 7)$$
Edit 2: A few worked examples to get familiar with the $\equiv$ notation.