For this type of questions you need to be familiar with definitions.

Here are the definitions you need:

Reflexive - A relation R on a set A is called reflexive if we have xRx for all x in A. That is,everyelement in A maps to (is related to) itself

Symmetric - A relation R on a set A is called symmetric if whenever xRy, then yRx for all x,y in A. That is, xRy => yRx. Note => means "implies" which is a logical operator.

Transitive - A relation R on a set A is called transitive if whenever xRy and yRz, then xRz for all x,y,z in A. That is, (xRy ^ yRz) => xRz. Note ^ here represents the logic symbol "and."

Equivalence Relation - A relation R on a set A is called an Equivalence Relation if R is reflexive, symmetric and transitive.

other things to note: iff and <=> mean "if and only if"

Now to answer your question.

(a) xEy (if and only if) x – y is an even integer

We show that E here is an equivalence relation.

Note that we have x - y being even <=> x,y are of the same parity, that is, they are both even or both odd at the same time. Thus E is reflexive, since every even number and every odd number is related to itself, that is, we have xEx for all x in Z.

Moreover, we have E symmetric. Since if x - y is even, then y - x is even, that is xEy => yEx.

Finally, we show that E is transitive. If x - y is even, and y - z is even, then x, y and z are of the same parity, and so x - z is even. That is, xEy and yEz => xEz.

Since E is reflexive, symmetric and transitive, E is an equivalence relation.

(b) xEy (if and only if) x – y is an odd integer

We show E is not an equivalence relation. Note that we have x - y odd <=> x and y are of opposite parities, that is, if x is even, y must be odd, and if x is odd, y must be even. Thus we do not have have any even number or any odd number related to itself, that is, we do not have xEx for any x in Z. This means E is not reflexive and hence cannot be an equivalence relation.