A={x∈Z:0≤x≤12}={0,1,2,3,4,5,6,7,8,9,10,11,12}R={(a,B):∣a−b∣is a multiple of 4}For any element a∈A, we have (a,a)∈R as ∣a−a∣=0 is a multiple of 4.∴R is reflexive.Now, let (a,b)∈R⇒∣a−b∣ is a multiple of 4.⇒∣−(a−b)∣=∣b−a∣ is a multiple of 4.⇒(b,a)∈R∴ is symmetric.Now, let (a,b),(b,c)∈R⇒∣a−b∣ is a multiple of 4 and ∣b−c∣ is a multiple of 4.⇒(a−b) is a multiple of 4 and (b−c) is a multiple of 4.⇒(a−c)=(a−b)+(b−c) is a multiple of 4.⇒∣a−c∣ is a multiple of 4.⇒(a,c)∈R.∴R is transitive.Hence, R is an equivalence relation.The SET of elements related to 1 is {1,5,9} since ∣1−1∣=0 is a multiple of 4,∣5−1∣=4 is a multiple of 4, and ∣9−1∣=8 is a multiple of 4Hence, R is an equivalence relation.(ii)A={x∈Z:0≤x≤12}={0,1,2,3,4,......,11,12}R={(a,b):a=b}={(0,),(1,1),(2,2),........,(11,11),(12,12)}For any element a∈A, we have (a,a)∈R, since a=a.∴R is reflexive.Now, let (a,b)∈R.⇒a=b⇒b=a⇒(b,a)∈R∴R is symmetric.Now, let (a,b)∈R and (b,c)∈R.⇒a=b and b=c⇒a=c⇒(a,c)∈R∴R is transitive.Hence, R is an equivalence relation.The elements in R that are related to 1 will be those elements from set A which are equal to 1.Hence, the set of elements related to 1 is {1}.