Reflexive, symmetric, transitive, antisymmetric, equivalence or partial order Let $X = \{0, 1, 2, 3, 4\}$ and  $\Delta X = \{(x,x)~:~x\in X\}$.
$$\Delta X = \{(0,0),(1,1),(2,2),(3,3),(4,4)\}$$
$R1 = ∆X∪ {(1,0), (0,1), (2,4), (4,2), (1,3), (3,1), (0,4), (4,0)}$
Question: Is R1 is reflexive, symmetric, transitive, antisymmetric, equivalence or a partial order? 
Reflective? No. a ∈ a such that (a, a) ∉ R1. 
Symmetric? No. 
Transitive? No. (a, b) ∈ R1 and (b, c) ∈ R1 do not imply (a, c) ∈ R1.
Antisymmetric: Yes.
Equivalence: No.
Partial Order? No.
Is this correct? 
 A: 
Reflective? No. a ∈ a such that (a, a) ∉ R1.

This is incorrect unless you can give a specific counter-example.  WHICH $a \in X$ is it, so that $(a,a) \not \in R1$.  $(0,0) \in R1$ so $0\in X$ is not the counter example.  So which $a \in X$ is the counterexample.

Symmetric? No.

This is incorrect unless you can give a specific counter-example.  WHICH $a,b \in X$ is it, so that $(a,b) \in R1$ but $(b,a)\not \in R1$.  Both $(0,1)$ and $(0,1)$ are in $R1$ so $0,1$ is not the counter example.  Neither $(0,2)$ nor $(2,0)$ are in $R1$ so $0,2$ is not the counter example.  Which $a,b$ is the counter example.

Transitive? No. (a, b) ∈ R1 and (b, c) ∈ R1 do not imply (a, c) ∈ R1.

Why not?  Which is your counter example?  $(2,2)$ and $(2,4)$ are in $R1$ and $(2,4)$ are in $R1$ so that is not a counter example.  Which counter example do you have?
Transitive? No. (a, b) ∈ R1 and (b, c) ∈ R1 do not imply (a, c) ∈ R1.

Antisymmetric: Yes.

Why?  Do you have a reason to state that if $(a,b) \in R1$ and $(b,a) \in R1$ that $a = b$.  Have tested them all?

Equivalence: No.

Why not.  What does equivalence mean?  Have you stated that equivalence means, reflexive, symmetric, and transitive and that this fails one of these?  Which ones?
You can't have a correct answer if you don't give reason.

Partial Order? No.

What does partial order mean and why doesn't this have it.  Have you stated that partial order means reflexive, anti-symmetric, and transitive?  Have you stated if $R1$ fails at least one of these?
If not, you have no reason.

Is this correct? 

No.  It isn't. 1) You give no reasons or counter examples so even if all your answer were correct, this would be incorrect.  2) Your answer are not all correct.  A few of them are but not all of them.  But since you gave no justification or counter-examples none of them can be accepted.
