As @AndrewHughes says in the comments . . . If $G$ is cyclic, then $G$ would contain all normal subgroups of orders dividing $|G|$. But $G$ does not contain a normal subgroup of order $q$. Hence $G$ is not cyclic.

Yes, it's okay. A generator $g$ of a group $G$ is an element that, potentially with other elements, generate the group $G$; which is to say that all other elements of $G$ are products of the generators.

Yes, this is correct. However, as @AlanWang points out, it would be even better if you declared that $a,b>0$.

As @arctictern put in a comment . . . Because $3^2\nmid 4!$, every $3$-sylow subgroup has order $3$, so they must be cyclic. But $4<3+3$, they must be generated by $3$-cycles. All $3$-cycles are conjugate.

As @DerekHolt states in the comments . . . Any diagonal matrix with entries from $\{-1,1\}$ and determinant $1$, excluding the identity, generates a subgroup of $\operatorname{SO}(6)$ isomorphic to $\Bbb Z_2$.