For First set A=x∈R:∣x−2∣>1
⇒ Value which A contains lies from x∈(−∞,1)∪(3,∞)
For second set B=x∈R:x2−3>1
⇒ Value which B contains lies from x∈(−∞,−2)∪(2,∞)
For third set C=x∈R:∣x−4∣⩾2
⇒ Value which C contains lies from x∈(−∞,2]∪[6,∞)
Now, (−∞,−2)∪[6,∞)∈(A∩B∩C)
Therefore, (A∩B∩C)c∩Z=−2,−1,0,1,2,3,4,5
Number of Subsets of above set =28=256