Wait after overcoming my fear of using $gH$ when what is meant is $g \mod H$, I think I figured it out. Might as well just type it here as an answer.
To prove $G = \langle g_1, \ldots, g_m\rangle H$:
$\supseteq$ duh
$\subseteq$ Let $g \in G.$ Then $gH = \pi_H(g) \in \pi_H(G)=\pi_H(\langle g_1, \ldots, g_m \rangle) = \{kH\mid k \in \langle g_1, \ldots, g_m \rangle \}$. Then $gH=kH$, for some $k \in \langle g_1, \ldots, g_m \rangle$. Hence, $gh=kl$, for some $h,l \in H$. Therefore, $g = \underbrace{k}_{\in \langle g_1, \ldots, g_m \rangle}\underbrace{lh^{-1}}_{\in H}$
Wait I also figured out $AH=BH$ (given $\pi_H(A) = \{a \bmod H\mid a \in A\} = \{b \bmod H\mid b \in B\} = \pi_H(B)$) even if you use $\bmod H$:
$\supseteq$ By symmetry, 1 direction is sufficient.
$\subseteq$ Let $ah \in AH$, for $(a,h) \in A \times H$. We must show that $ah=bk$, for some (b,k) \in B \times H.
Now, when $a \bmod H = \pi_H(a) \in \pi_H(A) = \pi_H(B) = \{b \bmod H\mid b \in B\}$. Then $a \bmod H = c \bmod H$, for some $c \in B$. Hence, $ca^{-1}=q$, for some $q \in H$. Therefore, $ah=cq^{-1}h$. Choose $(b,k)=(c,q^{-1}h)$. (or simply note that $ah=\underbrace{c}_{\in B}\underbrace{q^{-1}h}_{\in H} \in BH$.)
All you need to know is the first isomorphism theorem. This map is a surjection onto $G_{1}$.
So $(G_{1}\times G_{2})/\ker(\pi_{1})\cong G_{1}$.
If you want to proceed directly by hand then you can define $f:(G_{1}\times G_{2})/\ker(\pi_{1})\to G_{1}$ by $f((a,b)\ker(\pi_{1}))=a$. Now observe that the $\ker(f)=\{(e,b)\in G_{1}\times G_{2}:b\in G_{2}\}=\ker(\pi_{1})$. which is the identity element in $G_{1}\times G_{2}/\ker(\pi_{1})$. This gives you injectivity. And for any $a\in G_{1}$ . The equivalence class $(a,e)\ker(\pi_{1})\in (G_{1}\times G_{2})/\ker(\pi_{1})$ is the preimage. Hence done.
Note that the equivalence classes can be written as $(a,G_{2})\ker(\pi_{1})$ as only the first components give different equivalence class. $(a,b_{1})\ker(\pi_{1})=(a,b_{2})\ker(\pi_{1})\,\,,\forall b_{1},b_{2}\in G_{2}$. this is because $(a,b_{1})^{-1}\circ (a,b_{2})=(e,b_{1}^{-1}b_{2})\in \ker(\pi_{1})$.
Best Answer
To answer your first question, the group $\langle (123)(45)\rangle$ is a finite cyclic group, so it very much behaves like the cyclic subgroups of $\mathbb{Z}$, i.e. there is a minimal $k$ such that $((123)(45))^k=e$ after which the elements repeat.
Clearly we have $(123)^3=e$ and $(45)^2=e$, and moreover $(123)^{3m}=e$ and $(45)^{2n}=e$ for any two non-negative integers $m$ and $n$. We are therefore interested in those powers $3m$ and $2n$ for which the cycles $(123)$ and $(45)$ simultaneously go to $e$, i.e. in the minimal solution of $3m=2n$, which is $(m,n)=(2,3)$. In particular, we have $$((123)(45))^6=(123)^6(45)^6=(123)^{3\cdot 2}(45)^{2\cdot 3}=e$$ and therefore $$((123)(45))^{6+i}=((123)(45))^i$$ for any non-negative integer $i$.
Therefore to list the non-trivial elements of $\langle (123)(45)\rangle$ we have to figure out $$((123)(45))^i=(123)^i(45)^i$$ for $i=1,2,3,4,5$, which are $$(123)(45), (132), (45), (123),\text{ and }(132)(45).$$