I don't understand why $(1-i)$ is in the set of element of $R_2$ in the solution.
My understanding is that $\mathbb{Z}/2\mathbb{Z} = \{0,1\}$, endowed with addition and multiplication mod 2. So wouldn't the elements of $R_2$ be $\{0,1,i,1+i\}$? And shouldn't the multiplication table of $R_2$ be something like this?
Yes, you are right, the elements of $R_2$ are $0$, $1$, $i$, and $1+i$. And since, in $\mathbb{Z}/2\mathbb{Z}$, $1=-1$, $1+i=1-i$ there.