Well, in (1) I think I can replace $\mathbb{Z}$ by an arbitrary group $G$, because $\mathbb{Z}$ do not come in mind so quickly.

Sorry, my mistake, this is true for every group G, like Primoz just said. The assumptions are needed for it to say that $\pi_3 SX \cong ker(\kappa)$ for a space $X$, where $\kappa : \pi_1 X \otimes \pi_1 X \to (\pi_1 X)'$ is the commutator epimorphism. These are: $X$ is path-connected and $\pi_2 X = 0$. Then you can see that the above follows.