I understand G to be a group. I also know that by definition <a> is a subgroup of all the power of a. I know that the subgroup <a> has an order 0f 21  that is, it has 21 elements. I know that a group can be a subgroup of itself. Does that mean that G is the subgroup of itself? I am confused/confusing myself.

Yes, you are confusing yourself!
.
It does mean that <a> is G itself. That is G consists of the elements, a, a<sup>2</sup>, a<sup>3</sup>, a<sup>3</sup>, a<sup>4</sup>, ..., a<sup>21</sup> (and a<sup>21</sup>= e).
Okay, the subgroup generated by a<sup>7</sup> is, as you said, the subgroup of its powers: a<sup>7</sup>, (a<sup>7</sup>)<sup>2</sup>= a<sup>14</sup>, (a<sup>7</sup>)<sup>3</sup>= a<sup>21</sup>= u. How many members does that subgroup have?
Now do the same thing with a<sup>3</sup>. How many powers can you take until you get to a<sup>21</sup>= u?
xy= (a<sup>7</sup>)(a<sup>3</sup>)= a<sup>10</sup>. Same question as above.
I'll let you think longer about part ii.
Let G be a group, and let x be and element of G. Assume that x has infinite order. Prove that every nonzero power of x also has infinite order. That is, prove that, if i does not equal 0, then x<sup>i</sup> has infinite order. (Hint: This is a simple proof by contradiction, What happens if (x<sup>i</sup>)<sup>j</sup> = u for some positive integer j?)

Continuation of hint: (x<sup>i</sup>)<sup>j</sup>= x<sup>i+j</sup>!