Actually, $(0,1)+\langle(2,4)\rangle$ is not of order $4$. Note that
$$\langle (2,4)\rangle = \{ (2,4), (0,8), (2,2), (0,6), (2,0), (0,4), (2,8), (0,2), (2,6), (0,0)\}$$
so
$$\bigl( (0,1)+\langle (2,4)\rangle\bigr) + \bigl( (0,1)+\langle (2,4)\rangle\bigr) = (0,2)+\langle(2,4)\rangle = (0,0)+\langle(2,4)\rangle.$$
In fact, note that $\langle(2,4)\rangle = \langle (2,0), (0,2)\rangle$, which should make the isomorphism type of the quotient very clear.
For the second, notice that $(3,0)\in\langle(1,2)\rangle$. So you can first mod out by $(3,0)$ as a first approximation; we have $\mathbb{Z}\times\mathbb{Z}_6/\langle(3,0)\rangle \cong \mathbb{Z}_3\times\mathbb{Z}_6$. Now you want to quotient out this by the subgroup generated by (the image of) $(1,2)$.
$$\langle(1,2)\rangle = \{ (1,2), (2,4), (0,0)\}.$$
So $(0,1)+\langle (1,2)\rangle$ is of order $6$, which shows that the quotient is cyclic of order $6$.
Intuitively, taking the quotient modulo $\langle(1,2)\rangle$ "identifies" the $1$ in $\mathbb{Z}$ with the $2$ in $\mathbb{Z}_6$; that means that the $1$ in $\mathbb{Z}$ is of order $3$ in the quotient (as we saw), and that twice $(0,1)$ is the same as $(1,0)$. So from $(0,1)$ has order $6$ in the quotient; since $(1,0)$ and $(0,1)$ generate $\mathbb{Z}\times\mathbb{Z}_6$, knowing their images tells you exactly what happens to the whole group.
$\langle(1, 1, 1)\rangle$ is, as you said, isomorphic to $\mathbb{Z}_{40}$, and the prime factorization of $40$ is $5 \times 8$, so we can say that $\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5 \times \mathbb{Z}_8$. Now, it is important to convince yourself that such a group can't be expressed as the product of three subgroups, simply because $\mathbb{Z}_8$ can't be split since is cyclic and $\mathbb{Z}_5$ only has trivial subgroups, or, more formally, by contradiction, let $\langle(1, 1, 1)\rangle \simeq G_1 \times G_2 \times G_3 $. Since every subgroup of a cyclic group is cyclic, $G_2 \times G_3$ is cyclic and none of its coordinates is $\{e\}$. Therefore, $ G_2 \times G_3 $ is a product of cyclic groups whose orders are multiples of $2$ (so they're not coprime) and it's cyclic, which is a contradiction. As for how you could use that other method in this case, i'd say the most it can do for you is to simplify the thing as $(\mathbb{Z}_5 \times \mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1, 1)\rangle \simeq \mathbb{Z}_5/\langle1\rangle \times (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle \simeq (\mathbb{Z}_4 \times \mathbb{Z}_8)/\langle(1, 1)\rangle$, but since that $\langle(1, 1)\rangle$ is not a product of subgroups you have to figure out some other way to find it. I hope to have answered your question, let me know if that's not the case.
Best Answer
Proof: Let
$$\begin{align} \varphi: G\times H&\to (G/M)\times (H/N)\\ (g,h)&\mapsto (gM, hN). \end{align}$$
Clearly $\varphi$ is a well-defined, surjective homomorphism. (Why?)
Then $\ker(\varphi)=M\times N$.
Now, by the First Isomorphism Theorem,
$$(G\times H)/(M\times N)\cong (G/M)\times (H/N).\,\square$$