Find the amount of subgroups of order $3$ and $21$ in non-cyclic abelian group of order $63$.
In first case I found the amount of elements that have order $3$ - there are $8$ of them, in second case there are $48$ elements of order $21$. How do I connect these values with the amount of subgroups now?
By the Fundamental Theorem of Finitely Generated Abelian Groups, the only non-cyclic abelian group of order $63$ is (up to isomorphism) $\mathbb{Z}_{21} \times \mathbb{Z}_3$
You have successfully found that there are $8$ elements of order $3$, so now look at the subgroups generated by these particular elements. You will notice that you will find that some of these generated subgroups are the same. For example, the subgroup generated by $(7,1)$ is the same as the one generated by $(14,2)$
Use the same trick for the subgroups of order $21$