= Solution
The <Einstein tensor> is $G_{\alpha\beta}=R_{\alpha\beta}-\tfrac12g_{\alpha\beta}R$. Taking the <metric trace> of the <Einstein field equations> in four dimensions gives
$$
-R=8\pi T,
$$
and substitution back gives the <Trace-reversed Einstein field equations in D dimensions> specialized to $D=4$:
$$
\boxed{R_{\alpha\beta}=8\pi\left(T_{\alpha\beta}-\frac12g_{\alpha\beta}T\right)}.
$$
For <linearized gravity>, the given equation and $G_{\mu\nu}=8\pi T_{\mu\nu}$ imply
$$
\Box\bar h_{\mu\nu}=-16\pi T_{\mu\nu}.
$$
Since $\bar h=-h$, its trace is $\Box h=16\pi T$. Using $\bar h_{\mu\nu}=h_{\mu\nu}-\tfrac12\eta_{\mu\nu}h$ then yields
$$
\boxed{\Box h_{\mu\nu}=-16\pi\left(T_{\mu\nu}-\frac12\eta_{\mu\nu}T\right)}.
$$
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