See the Wikipedia article on finitely-generated abelian groups for more information. We have
Theorem A. Any finite abelian group $G$ admits a unique invariant factor decomposition.
Theorem B. Any finite abelian group $G$ admits a unique primary decomposition.
These decompositions are different, but logically Thm A $\Leftrightarrow$ Thm B: that is, once you prove one, the other follows from Sun-Ze (aka CRT), ${\bf Z}/nm\cong {\bf Z}/n\times{\bf Z}/m$ when $(n,m)=1$. Further one can compute one decomposition from the other (+ vice-versa). Let's discuss how.
Suppose we have the invariant factor decomposition of $G$ from Thm A as follows:
$$G\cong \frac{\bf Z}{n_1}\times\frac{\bf Z}{n_2}\times\cdots\times\frac{\bf Z}{n_l}.$$
Then each $n_i$ factors as $\prod_pp^{e(i,p)}$ for primes $p$ and exponents $e(i,p)$. By SZ then
$$G\cong\prod_{i=1}^l\frac{\bf Z}{n_l}\cong\prod_{i=1}^l\prod_p\frac{\bf Z}{p^{e(i,p)}}\cong\prod_p\left[\prod_{i=1}^l\frac{\bf Z}{p^{e(i,p)}}\right].$$
Notice the factors $\prod_{i=1}^l{\bf Z}/p^{e(i,p)}$ are $p$-groups; this is the primary decomposition, which is the same as the Sylow decomposition (as a direct product of Sylow $p$-subgroups, available for all nilpotent groups) when our group is abelian. It is probably best to think of Sylow theory as the inevitable noncommutative generalization of arithmetic in the theory of finitely-generated abelian groups.
Our original hypothesis that we had the invariant factor decomposition means that $n_i\mid n_{i+1}$, so we must have $e(i,p)\le e(i+1,p)$ for each $p$. Also, some of the $e$'s may be $0$. Thus if we begin with a primary decomposition as regarded in Thm B, we may order the prime exponents in an increasing order, and fill in extra $0$s as needed so that each prime has the same number of exponents, then form the $n_i$ out of the prime powers created from these exponents.
Here's an example. From a primary decomposition we obtain
$$\begin{array}{llccccc} G & \cong & \left(\frac{\bf Z}{2} \times \frac{\bf Z}{2}\times\frac{\bf Z}{8}\right) & \times & \left(\frac{\bf Z}{3}\times\frac{\bf Z}{27}\right) & \times & \left(\frac{\bf Z}{5}\right) \\ & \cong & \left(\color{Magenta}{\frac{\bf Z}{2}} \times \color{Blue}{\frac{\bf Z}{2}} \times \color{Green}{\frac{\bf Z}{8}} \right) & \times & \left(\color{Magenta}{\frac{\bf Z}{1}} \times \color{Blue}{\frac{\bf Z}{3}} \times \color{Green}{\frac{\bf Z}{27}} \right) & \times & \left(\color{Magenta}{\frac{\bf Z}{1}} \times \color{Blue}{\frac{\bf Z}{1}} \times \color{Green}{\frac{\bf Z}{5}} \right) \\ & \cong & \left(\color{Magenta}{\frac{\bf Z}{2}\times\frac{\bf Z}{1}\times\frac{\bf Z}{1}} \right) & \times & \left( \color{Blue}{\frac{\bf Z}{2}\times\frac{\bf Z}{3}\times\frac{\bf Z}{1}} \right) & \times & \left( \color{Green}{\frac{\bf Z}{8}\times\frac{\bf Z}{27}\times\frac{\bf Z}{5}} \right) \\ & \cong & \frac{\bf Z}{2\times1\times1} & \times & \frac{\bf Z}{2\times3\times1} & \times & \frac{\bf Z}{8\times27\times5} \\[7pt] & \cong & {\bf Z}/2 & \times & {\bf Z}/6 & \times & {\bf Z}/1080 \end{array}$$
as our invariant factor decomposition, with $2\mid6\mid1080$. And to see how to go in reverse with this same example, just read from bottom to top.
In conclusion, then, there are two standard fundamental representations of a finite abelian group, the invariant-factor decomposition and the primary decomposition. Theorems A and B state their existence and uniqueness; these are roughly logically equivalent using the remainder theorem, and it is relatively easy to move between the two decompositions using prime factorization. The IF representation invokes linear algebra, the primary decomposition invokes group theory (it is positioned within Sylow theory), and they are both positioned within commutative algebra.
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
1) This is a nontrivial theorem of Laszlo Lovasz. Google his name along with direct product and you will find it.
2) Let $$G = \mathbb{Z} / p \mathbb{Z} \times \mathbb{Z} / p \mathbb{Z} \times \ldots$$ $$H = \mathbb{Z} / p \mathbb{Z} $$ $$K = \mathbb{Z} / p \mathbb{Z} \times \mathbb{Z} / p \mathbb{Z}.$$