Given the set $A=\{1,2,,3,4,5\}$ and the relation $R=\{(a,b):|a-b| \;is\; even\}$:

Let $a=b$, $(a,a) \in R \rightarrow |a-a|=0$ which is even. Therefore $R$ is reflexive.

For $R$ to be symmetric, if $(a,b) \in R \rightarrow (b,a) \in R$.

$(a,b) \in R \rightarrow |a-b|=even$

$(b,a) \in R \rightarrow |b-a|=even$

$(a-b)=-(b-a); $ therefore $|(a-b)|=|(b-a)|$

Therefore $(b,a)\in R$. Hence, $R$ is symmetric

Let $(a,b) \in R \; and \;(b,c)\in R$

$\Rightarrow |a-b|\; is\; even$ and $|b-c|\;is \;even$

If $(a,c) \in R \rightarrow |a-c| = even$

Now, $a-c=a-b+b-c$, which is an even number, as the sum of even numbers is even.

Hence $R$ is transitive.