Zero divisors

(a) $\mathbb{Z}/6\mathbb{Z}$. $\bar2\cdot\bar3=\bar6=\bar0$, so $\bar2$ and $\bar3$ are zero divisors. Also $\bar4\cdot\bar3=\bar{12}=\bar0$. Zero divisors: $\bar2,\bar3,\bar4$.

(b) $\mathbb{Z}/7\mathbb{Z}$: $7$ prime, so every nonzero element is invertible, hence no zero divisors. No zero divisors (field).

(c) $M_2(\mathbb{R})$: $\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$, so non-invertible nonzero matrices are zero divisors.

Conclusion. (a) has zero divisors $2,3,4$; (b) has none; (c) has many (all non-invertible nonzero matrices).