Let $G=\Bbb Z_{30} \times \mathbb Z_{36}$. Let $G=\Bbb Z_{30}\times \mathbb Z_{36}$. 
Divide G to direct product of cyclic gropus from order exponent of prime. Than find all the fairs of generators of G. 
I thought about $36\times30=1080=2^3\times3^3\times5$. 
 A: Hint:
$$\Bbb{Z}_{mn}\cong\Bbb{Z}_m\times\Bbb{Z}_n\iff\gcd(m,n)=1$$

You have managed to write $$\Bbb Z_{30}\times\Bbb Z_{36}\cong\Bbb Z_2\times\Bbb Z_3\times\Bbb Z_5\times\Bbb Z_{4}\times\Bbb Z_{9}\tag{1}$$ Let's try to find the number of generators in easier cases. If we look at the group $\Bbb Z_3\times \Bbb Z_9$, we observe that no single element $(a,b)$ can generate it, since there are $3\cdot 9=27$ elements, and $9(a,b)=(0,0)$ (the highest order of any element is $9$, so we can make atmost $9$ different elements with $(a,b)$). This happens because $3\mid 9$.
On the other hand $\Bbb Z_3\times\Bbb Z_8$ can be generated by a single element, for instance $(1,1)$ or $(2,3)$,... Which can easily be verified since $\Bbb Z_{24}\cong\Bbb Z_3\times\Bbb Z_8$. This may suggest that $(1)$ requires two generators.
Let's denote this generators $$\begin{align}\alpha=(a,b,c,0,0)\\\beta=(0,0,0,d,e)\end{align}$$ How many possibilities are there for $a,b,c,d,e$?
Hint:

How many generators does $\Bbb Z_n$ have? $$$$ Let us suppose that $a$ is a generator of $\Bbb Z_n$, then $$a,2a,3a,\dots,na$$ are all distinct elements, in particular $na\equiv_n0$ and $ka\not\equiv_n0$ for $k=1,2,\dots,n-1$. This is the same as saying that the only way to have $n\mid ka$ is to have $n\mid k$. This is precisely the case when $\gcd(a,n)=1$, that is, $a$ generates $\Bbb Z_n$ when $a$ and $n$ have no common factor. The generators of $\Bbb Z_{10}$ for instance is $\{1,3,7,9\}$.

