`because` Both elements a and b of `{a,b}` are also the elements of the set `{b,c,a}`
Therefore, `{a,b} sub {b,c,a}`
`rArr` Given statement is true
(ii) `because {a,e} sub` {x : x is a vowel in English alphabet}
`:. {a,e} sub {a,e,i,o,u}`
`because` Both elements a and e of first set are also the elements of the second set
`:. {a,e} sub` {x : x is a vowel in English alphabet}
Therefore, given statement is true
(iii) Since `2 cancel(in){a,3,5}`
Therefore, the statement `{a} sub {a,b,c}` is true
(v) `{a} in {a,b,c}`
`because {a}` is not en element in `{a,b,c}`
Therefore, given statement is false.
(vi) `because` {x : x is an even natural number less than 6} `sub` {x : x is a natural number which divides 36}
`:. {2,4}sub{1,2,3,4,6,9,12,18,36}`
`:.` Both elements of first set are also in second set.
Therefore, given statement in true.