The fundamental theorem of finitely generated abelian groups does not say that you can split up $360$ in any way you want as a product of primes and obtain an isomorphism. It says that for any abelian group of order $360$, there is a way to write $360$ as a product of prime powers $360=p_1^{r_1}\cdot\cdots \cdot p_n^{r_n}$ such that $G\cong \mathbb Z_{p_1^{r_1}}\times Z_{p_n^{r_n}}$.
The appearance of the term "Betti number" in your theorem should be seen as a definition. The Betti number of a finitely generated abelian group $G$ is the number of $\mathbb{Z}$ factors appearing in the decomposition claimed by your theorem, or differently stated, the rank of $G$ as a $\mathbb{Z}$-module.
You can also understand it as a sort of analogue to the dimension of a vector space. Indeed, tensoring $G$ with $\mathbb{Q}$ will convert it into a $\mathbb{Q}$-vector space whose dimension is exactly the rank/Betti number of $G$.
That is, for example, if $G=\mathbb{Z}^n$, then $n$ is the Betti number.
Another example is $G=\mathbb{Z}_{(p)}$. There is no $\mathbb{Z}$ factor, so the Betti number is $0$.
Now let us look at your two statements:
The first one is false: Take $G=\mathbb{Z}$ and $H=\mathbb{Z}\times \mathbb{Z}_{(2)}$. They both have Betti number $1$, but they are clearly not isomorphic.
The second one is true: if the Betti number was positive, then the group would have a subgroup isomorphic to $\mathbb{Z}$, in particular it would have infinitely many elements.
Best Answer
We have
$\qquad \mathbb{Z}\times\mathbb{Z} = \mathbb{Z} e_1 \oplus \mathbb{Z} e_2 $
$\qquad \langle(0,3)\rangle = \mathbb{Z} (0 e_1) \oplus \mathbb{Z} (3e_2) $
Therefore,
$\qquad \mathbb{Z}\times\mathbb{Z}/\langle(0,3)\rangle \cong \mathbb{Z}\times\mathbb{Z_3}$
An explicit isomorphism is induced by $(x,y) \in \mathbb{Z}\times\mathbb{Z} \mapsto (x, y \bmod 3) \in \mathbb{Z}\times\mathbb{Z_3}$.