You have the right idea: $|\mathrm{Im}(\phi)|$ divides $|H|$, but on the other hand it is equal to $\frac{|G|}{|\ker(\phi)|}$, which is a divisor of $|G|$. Since $|H|$ and $|G|$ are coprime, this means that $|\mathrm{Im}(\phi)|=1$.
It should not be surprising that if a homomorphism $\phi$ is injective, then its kernel is trivial. After all, injectivity requires that the preimage of every element of the image be unique. (And the homomorphism property requires that the preimage of the identity be the identity, in particular.)
It is the converse -- that the triviality of the kernel is sufficient for injectivity -- that is less obvious. Mark Bennett has given the core idea: symbolically, if $\phi(a)$ equals $\phi(b)$, then their (group) inverses are equal too, and thus
$$\phi(ab^{-1})=\phi(a)\phi(b^{-1})=\phi(a)\phi(b)^{-1}=e$$
The first and second steps of this calculation work only because of the homomorphism property of $\phi$. Now, if the kernel is trivial, we conclude that $ab^{-1}=e$, so $a=b$, and injectivity follows.
Here is a nonstandard but alternative way to think about this result; it is good for intuition and illustrates another Big Idea. Say a function $f$ is injective at $y$, an element of the image, if the preimage $f^{-1}(y)$ has only one element. This is a "local" property; a function might be injective at $y_1$ but not at $y_2$.
For general functions between sets, injectivity at a point is not sufficient to deduce global injectivity -- that is, injectivity at every point. But if $f$ is a group homomorphism, not just a bare function between sets, then a special type of local injectivity, namely injectivity at the identity element, is sufficient to deduce global injectivity. The extra structure provided by the homomorphism property and the definition of the identity allow us to parlay a local phenomenon into a global one.
This is a taste of a general class of "local-to-global" results that show how local phenomena can be used to deduce global structure.
Best Answer
By the first isomorphism theorem, $A/\ker\varphi \cong \varphi(A)$. In particular: \begin{equation}\frac{29}{|\ker\varphi|}=|\varphi(A)|.\end{equation} Since 29 is a prime number, either $\ker\varphi=\{0\}$ (which implies $\varphi$ is injective) or $\ker\varphi=A$ (i.e. $\varphi$ is trivial).
If $|B|=80$ then since $\varphi(A)$ is a subgroup of $B$, $|\varphi(A)|$ divides $80$ by Lagrange's theorem. But $|\varphi(A)|$ also divides 29, so $|\varphi(A)|=1$ and $|\ker\varphi|=29$, which implies $\varphi$ is trivial.