Find all cosets of $H$ and $K$ in $G$ Let $G = \mathbb{Z}_{10} \times \mathbb{Z}_{4}$ and $H = \langle(3, 2) \rangle$ and $K = \langle(4, 2)\rangle$. Find all cosets to H and K.
We have 
\begin{equation*}
\begin{split}
G & = \mathbb{Z}_{10} \times \mathbb{Z}_{4} \\
& = \{0, 1, 2, \ldots, 9 \} \times \{0, 1, 2, 3 \} \\
& = \{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), \ldots, (1, 3), (2, 0), \ldots, (2, 3), \ldots, (9, 0), \ldots, (9, 3) \} ,
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
H & = \langle(3, 2)\rangle \\
& = \{k \cdot (3, 2) : k \in \mathbb{Z} \} \\
& = \{ (k \cdot 3, k \cdot 2) : k \in \mathbb{Z} \} \\
& = \{(0, 0), (3, 2), (6, 0), (9, 2), (2, 0), (5, 2), (8, 0), (1, 2), (4, 0), (7, 2) \} ,
\end{split}
\end{equation*}
since 
\begin{equation*}
\begin{split}
0 \cdot (3, 2) = (0, 0) \quad &\land \quad 1 \cdot (3, 2) = (3, 2) \quad \land \quad 2 \cdot (3, 2) = (6, 0) \\
3 \cdot (3, 2) = (9, 2) \quad & \land \quad 4 \cdot (3, 2) = (2, 0) \quad \land \quad 5 \cdot (3, 2) = (5, 2) \\
6 \cdot (3, 2) = (8, 0) \quad & \land \quad 7 \cdot (3, 2) = (1, 2) \quad \land \quad 8 \cdot (3, 2) = (4, 0) \\
9 \cdot (3, 2) = (7, 2) \quad & \land \quad \ldots 
\end{split}
\end{equation*}.
Now, in order to find all cosets of H, do I have to check for every $(a, b) \in G$ what sets I receive when I compute $(a, b) + H = \{(a, b) + h : h \in H \}$? I.e. do I have to compute 40 different sets? Is there a less brute force/tedious method?

Let me do it for $H$. You have found that $|H|=10$, so that there will be four cosets (by Lagrange).
One of them is $H$. You also know that distinct cosets are disjoint. Thus, pick an element not in $H$, say $(1,0)$; then a new coset is
$$
(1,0)+H=\{(1, 0), (4, 2), (7, 0), (0, 2), (3, 0), (6, 2), (9, 0), (2, 2), (5, 0), (8, 2) \}
$$
Is there an element not yet listed? Yes, $(0,1)$. Then a new coset is
$$
(0,1)+H=\{(0, 1), (3, 3), (6, 1), (9, 3), (2, 1), (5, 3), (8, 1), (1, 3), (4, 1), (7, 3) \}
$$
Now find an element not yet listed and do the same to find the fourth and last coset.
