(i) False
Let A={1,2} and B={1,{1,2},{3}}
2∈{1,2} and {1,2}∈{{3},1,{1,2}}
Now,
∴A∈B
However, 2/∈{{3},1,{1,2}}
(ii) False.
As A⊂B, B∈C
Let A={2},B={0,2}, and C={1,{0,2},3}
However, A/∈C
(iii) True
Let A⊂B and B⊂C.
Let x∈A
⇒x∈B [∵A⊂B]
⇒x∈C [∵B⊂C]
∴A⊂C
(iv) False
As, A/⊂B and B/⊂C
Let A={1,2},B={0,6,8}, and C={0,1,2,6,9}
However, A⊂C
(v) False
Let A={3,5,7} and B={3,4,6}
Now, 5∈A and A/⊂B
However, 5/∈B
(vi) True
Let A⊂B and x/∈B.
To show: x/∈A
If possible, suppose x∈A.
Then, x∈B, which is a contradiction as x/∈B
∴x/∈A