I think you just needed a reminder of the definition of $|$, but here are the ideas in a more spelled-out form than the comments had:
Showing $A_1\subseteq A_2$
For this part, we just have to show $\left(k∣bc+a \text{ and } k∣b\right)\Rightarrow k|a$, since the other condition for $A_2$, $k|b$, is included in the definition of $A_1$.
If $k∣bc+a$ and $k∣b$, then $(bc+a)/k$ and $b/k$ are integers, by the definition of "$|$". Multiplying the second of those by the integer $c$ tells us that $cb/k=bc/k$ is an integer, too.
But then $\frac{bc+a}k-\frac{bc}k=\frac{a}k$ is an integer because it's just a difference of integers.
Finally, $a/k$ being an integer means $k|a$, by definition.
Showing $A_2\subseteq A_1$
For this part, we just have to show $k|b\text{ and }k|a\Rightarrow k∣bc+a$, because $k|b$ is included in the defintion of $A_2$.
If $k∣b$ and $k∣a$, then $b/k$ and $a/k$ are integers, so "$c$ times the first plus the second" is an integer, too. But that number is $(bc+a)/k$, so $k|(bc+a)$.
Conclusion
Since $A_1\subseteq A_2$ and $A_2\subseteq A_1$, $A_1=A_2$. (If you had an element in one set but not the other to contradict the equality, it would contradict one of the two subset statements, too.)