I know that $p: \mathbb{R}\to \mathbb{S}^1, t\mapsto(\cos2\pi t, \sin2\pi t)$ is a covering space and so $p\times p:\mathbb{R}\times\mathbb{R}\to \mathbb{S}^1\times \mathbb{S}^1$ is also, and since $\mathbb{R}^2$ is simply connected then the subgroup associated with this covering space is trivial, then I already have an answer for (c).
I know that $Id\times p:\mathbb{S}^1\times\mathbb{R}\to \mathbb{S}^1\times \mathbb{S}^1$ and $p\times Id:\mathbb{R}\times\mathbb{S}^1\to \mathbb{S}^1\times \mathbb{S}^1$ and $Id\times Id:\mathbb{S}^1\times\mathbb{S}^1\to \mathbb{S}^1\times \mathbb{S}^1$ are also cover spaces but I do not know which subgroups of $\mathbb{Z}\times\mathbb{Z}$ correspond, could someone help me please?
(a) The subgroup of $\mathbb{Z}\times\mathbb{Z}$ generated by $m\times 0$ is $m\mathbb{Z}\times\{0\}$? To what covering space does this correspond?
(b) What is this subgroup of $\mathbb{Z}\times\mathbb{Z}$?
You are being asked to describe covering spaces associated to the subgroups $m\mathbb{Z} \times n\mathbb{Z}$, where $\pi_1(S^1) \cong \mathbb{Z}$ via $[\alpha] \mapsto 1$, where $\alpha(t) = (\cos{2\pi t}, \sin{2\pi t}$, $0 \leq t \leq 1$ and so $\pi_1(S^1 \times S^1) \cong \mathbb{Z} \times \mathbb{Z}$.
You can solve this problem by first describing a covering space of $S^1$ corresponding to $m\mathbb{Z} < \pi_1(S^1)$. What this means is that you are to find a covering $p:X \to S^1$ such that $p_*\pi(X) = m\mathbb{Z}$.
Here is a hint:
Since $m\mathbb{Z}$ has index $m$ in $\mathbb{Z}$, the covering must be $m$-sheeted. What is $X$? What is $p$? Think about what you know about coverings of the circle.