I gave a similar solution here.
For brevity, I'll just give a general idea: first, you should determine the exact contents of H. You can do this just by repeatedly exponentiating a10 until you get e. You should see that H={a10, a5, e}.
Then, you go through elements of G and find what coset they produce. You already know that a10, a5, and e are in < a10>. You should look at elements that you haven't yet seen in a coset, and see what you get. For example, a*< a10>={a11, a6, a}. Just keep repeating until every element has been put in a coset. Note that cosets partition the group, so you should see each element once and only once, and that each coset should be the same size- that of the subgroup, so here 3- so you should get 5 cosets total since there are 15 elements in the whole group.
I'll do the first one for you.
$H=<(123)>=\{(1),(123),(132)\}$. A left coset of $H$ is a set of the form $aH$, where $a\in A_{4}$. One thing you probably learned is that two left cosets are equal to one another or disjoint. Note that
$$
\begin{aligned}
&(12)(34)H=\{(12)(34), (243),(143)\}\\
&(13)(24)H=\{(13)(24), (142), (234)\}\\
&(14)(23)H=\{(14)(23), (134), (124)\}.
\end{aligned}
$$
Thus,
$$
A_{4}=H\sqcup(12)(34)H\sqcup(13)(24)H\sqcup(14)(23)H,
$$
where $\sqcup$ denotes a disjoint union.
Similarly, for right cosets, we have
\begin{aligned}
&H(12)(34)=\{(12)(34), (134),(234)\}\\
&H(13)(24)=\{(13)(24), (243), (124)\}\\
&H(14)(23)=\{(14)(23), (142), (143)\}.
\end{aligned}
Thus,
$$
A_{4}=H\sqcup H(12)(34)\sqcup H(13)(24)\sqcup H(14)(23).
$$
Now, note that $(12)(34)H\not=H(12)(34)$. Thus, $H$ is not normal in $A_{4}$.
Recall that when computing cosets, the representative is not unique. That is, I used $(12)(34),(13)(24), (14)(23)$ and $(1)$ as the coset representatives. The idea is you start with $H$. Then, to get a different coset, choose $a\in A_{4}\setminus H$ to get $aH$. Then, to get another coset, choose $b\in A_{4}\setminus(H\cup aH)$, and so on, until you have all of the elements in $A_{4}$ contained in one of the cosets.
Best Answer
Let $\;z,w\in\Bbb C^*\;$ . Observe that
$$zT=wT\iff z^{-1}w\in T\iff\left|z^{-1}w\right|=1\iff|z|=|w|$$
Can you now see, both geometrically and algebraically, what the cosets in the quotient group $\;\Bbb C^*/T\;$ are?