In (i), the set , you have to look at the set of all powers of 3, where n is a (positive or negative) integer. If n is positive, will always be positive, but it can be arbitrarily close to 0. So 0 is a lower bound, but is never actually equal to 0. Therefore the g.l.b. is 0, but it is not in the set. If n is negative, will be a positive power of 3, and it can be arbitrarily large, so there is no upper bound to this set.
For (ii), the set is . So x runs through the real numbers, and you'll need to use calculus to check that this function has a maximum value 1/4 when x=1, and a minimum value –1/4 when x=–1. The function tends to 0 as and it never takes values outside the range from –1/4 to +1/4. So the l.u.b. is 1/4, the g.l.b. is –1/4, and both of those bounds lie in the set (because the function actually attains both of its extreme values).
In (iii), the set is . The best way to tackle this is probably just to write down the first few terms of the sequence and see what they look like. For n=1,2,3,4 you get the values 1, 4/5, 6/8, 8/11. These numbers are decreasing, and you'll need to convince yourself that the sequence continues to decrease, with limiting value 2/3 as . The initial value 1 is the l.u.b., and it is in the set. The g.l.b. is 2/3, and that value is never actually attained so it is not in the set.
Finally, in (iv), a real number x satisfies the condition if , so is the g.l.b and is the l.u.b. But we are told that , and the numbers are not rational. So neither the g.l.b. nor the l.u.b. is in the set in this case.
I hope that illustrates the sort of reasoning that you need to use to answer problems of this sort.