= Solution
\b[Compactifications.] A <Hausdorff compactification> of $X$ is a compact <Hausdorff space> $K$ together with a <topological embedding> $j:X\to K$ having dense image. We identify $X$ with that image.
For noncompact locally compact Hausdorff $X$, adjoin $\infty$ and give $X^+=X\cup\{\infty\}$ its <one-point compactification> topology: retain the open sets of $X$, and take $\{\infty\}\cup(X\setminus C)$, with $C$ compact, as a neighborhood base at $\infty$. Compact subsets are closed, making this a topology with the original <subspace topology> on $X$.
An open cover contains a neighborhood of $\infty$, leaving a compact subset of $X$ to cover by finitely many more members. Hence $X^+$ is compact. Points of $X$ are separated using its Hausdorff property; to separate $x\in X$ from $\infty$, use local compactness to choose an open $U\ni x$ with compact closure and take $X^+\setminus\overline U$ at $\infty$. Finally every neighborhood of $\infty$ meets $X$, because $X$ itself is not compact. Thus $X$ is dense and \b[$X^+$ is a <Hausdorff compactification> with one-point remainder].
\b[Comparison maps.] The <Stone-Čech compactification> of a <Tychonoff space> has the universal property that every <continuous map> from $X$ to a compact <Hausdorff space> extends continuously to $\beta X$. Applying it to any <Hausdorff compactification> $K$ gives a continuous surjection $\pi:\beta X\to K$ fixing $X$. Surjectivity follows because its compact image is closed and contains the dense set $X$.
We need the <fibers of compactification comparison maps> property
$$
\boxed{\pi^{-1}(X)=X.}
$$
Suppose $\pi(z)=x\in X$ with $z\ne x$. By regularity of compact Hausdorff $\beta X$, choose an <open neighborhood> $U$ of $x$ with $z\notin\overline U$. The embedding $X\subseteq K$ supplies open $V\subseteq K$ containing $x$ and satisfying $V\cap X\subseteq U\cap X$. The open set $\pi^{-1}(V)$ contains $z$. Density of $X$ implies $z\in\overline{\pi^{-1}(V)\cap X}=\overline{V\cap X}\subseteq\overline U$, a contradiction. Thus the displayed identity holds.
\b[Equivalence of (a), (b), (c).] Clearly (b) implies (c). If (c) holds in $K$, inverse images under $\pi$ and $\pi^{-1}(X)=X$ show that $X$ is a <G-delta set> in $\beta X$, giving (a). Conversely assume (a) and write $X=\bigcap_n U_n$ in $\beta X$. The sets $F_n=\beta X\setminus U_n$ are compact and miss $X$. Their images $\pi(F_n)$ are closed in $K$ and miss $X$ by the fiber identity. Since $\pi$ is onto, they exhaust $K\setminus X$. Hence
$$
\boxed{X=\bigcap_n\bigl(K\setminus\pi(F_n)\bigr).}
$$
This proves (b), for every $K$. This <G-delta compactification invariance> defines a <Čech-complete space>.
\b[The Baire property.] Suppose $X=\bigcap_n G_n$ in a compact Hausdorff $K$, with $X$ dense and all $G_n$ open. Each $G_n$ is dense. For dense open subsets $U_n$ of $X$, choose open $O_n\subseteq K$ with $U_n=X\cap O_n$. Every nonempty open subset of $K$ meets $X$ in a nonempty relatively open set and hence meets $U_n$, so each $O_n$ is dense in $K$.
The assumed <Baire category theorem> in $K$ makes $\bigcap_n(G_n\cap O_n)$ dense in $K$. This set lies in $X$ and equals $\bigcap_n U_n$, so it is also dense in $X$. Therefore \b[<Čech-complete spaces are Baire spaces>].
Enumerate $\mathbb Q=\{q_n:n\in\omega\}$. Each $\mathbb Q\setminus\{q_n\}$ is open and dense in $\mathbb Q$, while their intersection is empty. Thus $\mathbb Q$ is not a <Baire space>, and consequently
$$
\boxed{\mathbb Q\text{ is not a }G_\delta\text{ in any Hausdorff compactification of }\mathbb Q.}
$$
If $\mathbb Q$ were a <G-delta set> in $\mathbb R$, it would also be one in $\mathbb R^+$: the subspace $\mathbb R$ is open in its <one-point compactification>, and adjoining that open condition to a countable intersection preserves the <G-delta set> property. Since $\mathbb Q$ is dense in $\mathbb R$ and hence in $\mathbb R^+$, this would be a <Hausdorff compactification> of $\mathbb Q$ contradicting the preceding conclusion. Therefore $\boxed{\mathbb Q\text{ is not }G_\delta\text{ in }\mathbb R}$.
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