The answer to Question 1 is no.
Let $A=\mathbb{Z}/8\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ and let $B$ be the subgroup generated by $(2,1)$.
Since $B$ is cyclic of order $4$, if it were contained in a proper direct summand of $A$ then it would be contained in a cyclic subgroup of $A$ of order $8$, and so $(2,1)$ would be equal to $2a$ for some $a\in A$, which it's not.