1) Why $\mathbb Z/2\mathbb Z$ is not free as a $\mathbb Z$-module ? Indeed, let $1=1\cdot 1$ and thus $\{1\}$ is a basis, no ?
2) Does $\mathbb Z$ is a free $\mathbb Z$-module ? I would say yes with basis $\{1\}$, but since $\mathbb Z/2\mathbb Z$ is not free, I have doubt.
On
It is true that $\{1\}$ is a generating set (or a set of generators, if you prefer that phrase), but it is not a basis. In order to be a basis, every element in the module must be given by a unique linear combination of basis elements, and that's not the case here. For instance, $1\cdot1$ and $3\cdot1$ are two different linear combinations that give the same element in $\Bbb Z/2\Bbb Z$.
$\Bbb Z$, on the other hand, is a free $\Bbb Z$-module. That's because $\{1\}$ is a basis in this case. Indeed, if $n\cdot1=m\cdot1$, then $n=m$.
On
My addition, there is a universal property for free objects.
Let $S$ mark the set that generates the free object $F$ in question, then there is a set function (not homomorphism) $\imath:S\to F$. For this free object it is such that if we have a set function $f:S\to M$, where $M$ is another algebraic object of the desired type, then there exist a homomorphism $g$ such that $g\circ\imath = f$. For $\Bbb Z_2$ we have, as you said, $S=\{1\}$, now let $M=\Bbb Z$, then if $\Bbb Z_2$ is free we should have a homomorphism $\varphi$ such that it commutes, however the only homomorphism between $\Bbb Z_2$ and $\Bbb Z$ is the zero homomorphism, as such we have $\varphi\circ\imath(1)=0$ but $f(1)=1$ and as such it cannot be free.
For the other it follows equivalently.
1) No: $1$ is not free (linearly independent), since $2\cdot 1=0$.
2): Yes: $\mathbf Z$ is a free $\mathbf Z$-module of rank one.