$R =\{ (A,B) \ |\ A,B\ \epsilon \ P(\mathbb{N}), (A=B) \vee (A\cup\ \{ 1,2\} =B) \}$
What is the meaning of this relation? it means either of $A=B$ and $(A\cup\ { 1,2} =B)$ statements must be true?
hence it's a reflexive relation because the $A=B$ is true but the other not?
Right, if $A$ and $B$ are two subsets of $\mathbb N$ the the set $A$ relates to $B$ if either $A = B$ or $A$ would equal $B$ if you add $1$ and $2$ to $A$.
So $\{1,5,7,8\}R \{1,5,7,8\}$ and $\{1,5,7,8\}R \{1,2,5,7,8\}$ but $\{1,2,5,7,8\} \not R \{1,5,7,8\}$ for example.
It's weird relation but it's certainly a well defined relation.
It's reflexive as for all $A\subset \mathbb N$ we always have $A=A$ so $A \mathbb R$.
It's certainly not symmetric. If $\{1,2\}\subset B$ and $A = B\setminus\{1,2\}$ then $A\cup \{1,2\}=B$ so $ARB$ but $B\ne A$ and $B\cup\{1,2\} = B\ne A$ so $B\not R A$. (I gave a counter example in paragraph 2, but the simplest would be $\emptyset R\{1,2\}$ but $\{1,2\}\not R \emptyset$.)
I think it's anti-symmetric. If $A R B$ and $A\ne B$, then $A\ne A\cup \{1,2\}=B$. So $\{1,2\}\not \subset A$ but $\{1,2\}\subset B$ so $B\ne A$ and $B\cup\{1,2\}=B\ne A$ so $B\not A$. So $ARB$ and $BRA$ if only if $A=B$.
And I think it is Transitive.
If $A R B$ and $B RC$ then
case 1: $A = B$ than $BRC \implies ARC$.
case 2: $A\ne B$ then $A \cup \{1,2\}= B$. And $\{1,2\}\subset B$ so $B=B\cup \{1,2\}$. So $B RC$ means $B = C$ or $B\cup\{1,2\}=C$ but as $B= B\cup\{1,2\}$ it follows that $B=C$. So $ARB\implies ARC$.
So either way $ARB, BRC\implies ARC$.