1) A regular local ring has finite global dimension. Hence all modules have finite projective dimension, i.e. they have a finite projective resolution. Furthermore, on a local ring, finitely generated projective modules are free.

2) This is because $R[x^{-1}]_\mathfrak n=R_{\mathfrak n\cap R}[x^{-1}]_\mathfrak n$, and $R_{\mathfrak n\cap R}[x^{-1}]=R_{\mathfrak n\cap R}$ since, as $x\notin \mathfrak n\cap R$, $x$ is a unit in $R_{\mathfrak n\cap R}$.

The (Krull) dimension is no decreasing by localisation, and indeed it decreases here because $\mathfrak mR_{\mathfrak n\cap R}=R_{\mathfrak n\cap R}$.

3) $C_\mathfrak n$ is $0$ because $Q$ has finite presentation, so
$$\DeclareMathOperator{\Hom}{Hom}
\bigl(\Hom_{R[x^{-1}]}(Q,M)\bigr)_\mathfrak n\simeq\Hom_{R[x^{-1}]_\mathfrak n}(Q_\mathfrak n,M_\mathfrak n)\simeq M_\mathfrak n$$
since $Q_\mathfrak n$ is free of rank $1$. Thus $C_\mathfrak n$ is isomorphic to the cokernel of $M_\mathfrak n\longrightarrow N_\mathfrak n$, and localisation is exactâ€¦

This proves $C=0$, in other words the functor $\Hom_{R[x^{-1}]}(Q,\ \cdot)$ is right exact, which is the definition of $Q$ being a $R[x^{-1}]$ projective module.