As the title says I have: $G$ cyclic group of order 24, and $H= \langle x^6 \rangle$ , we also know that $x$ is a generator of $G$ so, $\langle x \rangle = G$
i have to find the order of each element of the quotient group $G/H$
and say how many generators have $G/H$
(answer adapted to some of the comments)
The order of $H$ is the order of its generator $x^6,$ i.e., 4. Therefore the order of $G/H$ is 6: there are 6 different cosets of the form $x^iH.$
We can choose the first powers of $x$ as representatives of these cosets:
$$G/H=\left\{H,xH,x^2H,x^3H,x^4H,x^5H\right\}$$
The order of an individual coset $x^iH$ is the smallest nonzero natural number $n$ such that $i.n$ is a multiple of 6.
Now at least one of the elements of $G/H$ turns out to have its order equal to the order of $G/H$ itself, implying that $G/H$ is cyclic.
A cyclic group is generated by an individual element $g$ iff the order of that element is the order of the group.