Lagrange states that for a finite subgroup $H$ of a group $G$: $|H|$ divides $|G|$
Converse of Lagrange: Given a group $G$ of finite order such that $|G| = n$, then for every divisor $d$ of $n$, there exists a subgroup $H\le G$ of order $d$.
1) Fundamental Theorem of finitely generated abelian groups: Any abelian group of order $n$ is isomorphic to direct product of cyclic groups (with the product of the orders of those cyclic groups = $n$.) It is possible to obtain subgroups of order $d$ (divisors of $n$) by taking appropriate divisors of the respective subgroups. You already know that the converse of Lagrange's Theorem is true for finite cyclic groups. As for infinite cyclic groups (all isomorphic to $\mathbb{Z}$), the only subgroup of finite order is the identity $\{0\}$ with order $1$, which divides all $n \in \mathbb{Z}$, though one cannot really divide the order of an infinite set. At any rate, Lagrange really only applies to finite groups of order $n$, where $n$ is fixed. It follows that the *converse holds for any [finite] abelian group.
(2) Suppose $G$ is of order $8$. Let $1\ne g\in G$. If $\langle g\rangle = G$, then $G$ is cyclic and we are done. Else, if the group has an element of order $4$, it has an element of order $2$ and we are done. (There are only two groups, up to isomorphism, of order $4$, and both have subgroups of order $2$.) Otherwise, we are left with a group of order $8$ such that each element of $G$ has order of at most 2, which means that the group is abelian...(why?). So we can conclude the converse holds for every group of order $8$.
(3) See this link. What are all the factors of $|S_4| = 24$? You'll see there exists an $H_i\le S_4$ such that for each factor $k_i$ of 24, there is a subgroup $H_i$ such that $|H_i| = k_i$. Check-mark: Converse of Lagrange holds for $S_4$.
(4) $A_4 \le S_4$ has order $12$. But: $A_4$ has no subgroup of order $6$. Why not? Since $6 \mid 12$, and there is no $H\le A_4$ such that $|H| = 6$, the converse of Lagrange fails in the case of $A_4$.