first of all is the group of units of the integers modulo 21 (all elements of that have a multiplicative inverse with respect to multiplcation modulo 21).

it turns out, that these are precisely the integers between 1 and 20 that are relatively prime to 21 (or more precisely, the equivalence classes of these integers mod 21).

so

which of these numbers are congruent to 1 mod 3?

clearly that set is {1,4,10,13,16,19} (note that this set has order 6, which divides 12).

which of these numbers are congruent to 1 mod 7?

these: {1,8}. note that this set also has order a divisor of 12.

i'll show that {1,8} is a cyclic group of order 2, under multiplication mod 21. to do this, it suffices to show that 8 is of order 2:

8^2 = 8*8 = 64 = 1 (mod 21) (since 64 = 63 + 1 = 21(3) + 1).

if {1,4,10,13,16,19} is indeed a subgroup of , then it is abelian (since multiplication mod 21 is commutatitive), and as such it must be cyclic of order 6. find a generator, and you're done!