I only answer the first part of the question. As it is well-known, $\chi$ and $k$ have the same parity. In particular, if $\chi$ is trivial then $k$ must be even. Conversely, if $\chi$ is non trivial then $f$ has CM by its own Nebentypus $\chi$ because $f$ is assumed to have rational (hence real) coefficients. Therefore $\chi$ is the quadratic character associated with an imaginary quadratic field $K$ of (negative) discriminant $D$ (see Ribet, *Galois representations attached to eigenforms with Nebentypus*, Modular Functions of One Variable V, LNM Vol. 601). In particular, we have $\chi(-1)=-1$ and $k$ is odd.

Let us now assume that we are in that latter case, i.e. $k$ is odd. By a result of Hecke and Shimura (*loc. cit.* pp. 34-36), $f$ is the newform attached to a Hecke character $\psi$ of $K$ with conductor, say, $\mathfrak{m}$ and infinite type $k-1$. (That is, if we view $\psi$ as a homomorphism from the group of prime-to-$\mathfrak{m}$ fractional ideals of $K$ into $\mathbf{C}^*$, then $\psi(\alpha \mathcal{O}_K)=\alpha^{k-1}$ for all $\alpha\in K^*$ with $\alpha\equiv 1\pmod{\mathfrak{m}}$; here $\mathcal{O}_K$ denotes the integer ring of $K$.) Moreover, in the notation of the question, we have $N=-\mathcal{N}(\mathfrak{m})D$ where $\mathcal{N}$ denotes the norm map from $K$ to $\mathbf{Q}$.

If $\mathfrak{p}$ is a prime of $\mathcal{O}_K$, denote by $e_{\mathfrak{p}}$ the exponent of the conductor $\mathfrak{m}$ of $\psi$ at $\mathfrak{p}$. By Proposition 6.1 and Table 1 of Schütt's paper *CM newforms with rational coefficients* (Ramanujan Journal **19** (2009), 187-205), for all prime ideals $\mathfrak{p}$, we have $e_{\mathfrak{p}}=e_{\overline{\mathfrak{p}}}$ and $e_{\mathfrak{p}}$ even if $\mathfrak{p}$ ramifies in $K$. (For a more detailed proof of these results, see chapter II of the author's dissertation available on his homepage.) This in turn implies that $\mathcal{N}(\mathfrak{m})=N/(-D)$ is a square, as desired.