Well, ${\zeta_8}^2$ is a fourth root of unity, and namely it is $i$ (at $\pi/2$), prefer to denote it $\zeta_8=\sqrt i$. This field $\Bbb Q(\sqrt i)$ is then a 4d vectorspace over $\Bbb Q$ with a standard basis $1,\sqrt i,\,i,\,i\sqrt i$. (The next power, ${\zeta_8}^4= -1$ is already dependent on them. Note also that $\Bbb Q(\sqrt i)=\Bbb Q(i,\sqrt 2)$ as field extension.) So that $\Bbb Z[\sqrt i] =\{a+b\sqrt i+ci+di\sqrt i \mid a,b,c,d\in\Bbb Z\}$.
- What are the integers in $\Bbb Q(\sqrt i)$? It is going to be $\Bbb Z[\sqrt i]$, but it needs to be thought over.
- As David mentioned in a comment, $2\in\mathscr P=\left((1-\sqrt i)\right)$, because (writing '$a\equiv b \pmod{\mathscr P}$' for $b-a\in\mathscr P$), we have $1\equiv\sqrt i \overset{()^2}\implies 1\equiv i \overset{()^2}\implies 1\equiv -1 \overset{+1}\implies 2\equiv 0 \pmod{\mathscr P} $
- Then, for $\mathscr P^2$, it is the ideal generated by $(\sqrt i -1)^2=i-2\sqrt i+1$. So, basically the relation $ 1+i \equiv 2\sqrt i \pmod{\mathscr P^2}$ generates it. From this, we also have $2i\equiv 4i $, implying $0\equiv 2i$ then (multplying by $(-i)$:) $\ 0\equiv 2 \pmod{\mathscr P^2}$, so again $-1\equiv 1$. Also, $1+i\equiv 2\sqrt i\equiv 0\sqrt i=0$, that is, $1\equiv -i=(-1)i\equiv i$. And, because $\sqrt i-1\notin\mathscr P^2$, we also have $\sqrt i\not\equiv 1 \pmod{\mathscr P^2}$. So, finally, we can conclude that $R/\mathscr P^2$ is represented by the following set: $\{ 0,1,\sqrt i,1-\sqrt i \}$ And, it is not $\Bbb Z/4\Bbb Z$, but still a ring with 4 elements..
I'm sure there is a more sophisticated and simpler solution, but until we find it, you can play around with $\sqrt i$..
About the other statement, $\mathscr P^{-2}/R$, well.. the problem is that $R$ cannot be an ideal in there, because $1\in R$.
Edit: But, as Hurkyl noted, they are both $R$-modules, so it can make sense anyway.