Let $G$ and $H$ be groups so that $|G|=21$, $|H|=49$. If $G$ has no normal subgroups of order 3, find all homomorphisms $f \colon G \to H$.

The kernel of $f\colon G\to H$ must be normal and hence cannot be of order $3$. Nor can it be of order $7$ or $1$ because then the image would have an element of order $3$. Hence $\ker f=G$.

Whenever you define a homomorphism $f \colon G \rightarrow H$, you will get the normal subgroup $\text{ker}(f) \subset G$. By Lagrange's theorem the order of a subgroup divides the order of group, such that the order of the kernel is either $1$, $3$, $7$ or $21$. By your assumption we can exclude $3$.

Now what happens in the other cases? You need to use Lagrange's theorem for the image (look at the order of the elements there as primes do not have that many divisors).

The image of $f$ is subgroup of $H$ and its order divides the order of $G$, being the index of the kernel of $f$. Therefore, the order of the image of $f$ divides $\gcd(21,49)=7$ and so is either $1$ or $7$. Since $G$ has no normal subgroups of order $3$ (that is, of index $7$), the image must have order $1$. Thus, $f$ is the trivial homomorphism.