I assume you are familiar with this lemma (or can prove it): If $n$ and $m$ are coprime, then $Z_{nm}=Z_n\times Z_m$ (where equality stands for isomorphism).
Using this lemma you can decompose all of your groups to products of cyclic groups of prime power order.
For example,
\begin{eqnarray}
Z_{33}\times Z_{15}\times Z_{15}
&=&
Z_{3\cdot11}\times Z_{3\cdot5}\times Z_{3\cdot5}
\\&=&
Z_3\times Z_{11}\times Z_3\times Z_5\times Z_3\times Z_5.
\end{eqnarray}
From here you can recognize that this group has a noncyclic subgroup $Z_3\times Z_3$, so the group itself is not cyclic.
A product of cyclic groups of coprime orders is cyclic, as is easy to check.
With these tools you can answer question 2.
To answer question 1, you can use a converse of the lemma I stated at the beginning.
But there is also a more direct way now that you have explicit groups.
If two groups have the same decomposition (as in your last question), they are isomorphic.
But for example $Z_5\times Z_5\neq Z_{25}$ because only one of these groups has an element of order 25.
Can you solve your problem and check your results with these ideas?
Addendum:
With the method I described you get
\begin{eqnarray}
Z_{33}\times Z_{15}\times Z_{15}
&=&
Z_3\times Z_{11}\times Z_3\times Z_5\times Z_3\times Z_5,\\
Z_{25}\times Z_{297}
&=&
Z_{25}\times Z_{27}\times Z_{11},\\
Z_{45}\times Z_{165}
&=&
Z_5\times Z_9\times Z_3\times Z_5\times Z_{11}\quad\text{and}\\
Z_{55}\times Z_9\times Z_{15}
&=&
Z_5\times Z_{11}\times Z_9\times Z_3\times Z_5.
\end{eqnarray}
Using the ideas I gave above, one can see that the second one is the only cyclic group, so others are not isomorphic to it.
Compare the decompositions of the last two groups: What do you notice?
Compare the first and the third group: Which one of these has elements of order 9?