= Solution
For this question use the <Levi-Civita connection>, its <metric compatibility> and the usual <curvature> symmetries. In the index ordering of Question 1, put $R_{\mu\nu\rho\sigma}=g_{\sigma\lambda}R_{\mu\nu\rho}{}^\lambda$ and $R_{\nu\rho}=g^{\mu\sigma}R_{\mu\nu\rho\sigma}$. Contract the <second Bianchi identity> in its derivative index $\mu$ and <curvature> index $\rho$. The <curvature> symmetries give
$$
\nabla^\rho R_{\nu\lambda\rho\sigma}
=\nabla_\lambda R_{\nu\sigma}-\nabla_\nu R_{\lambda\sigma}.
$$
Now contract $\lambda,\sigma$. The left-hand side becomes $-\nabla^\rho R_{\nu\rho}$, while the right-hand side becomes $\nabla^\sigma R_{\nu\sigma}-\nabla_\nu R$. Thus
$$
2\nabla^\mu R_{\mu\nu}=\nabla_\nu R.
$$
Using $\nabla g=0$ yields the <contracted Bianchi identity>
$$
\boxed{\nabla^\mu G_{\mu\nu}=0,\qquad
G_{\mu\nu}=R_{\mu\nu}-\tfrac12g_{\mu\nu}R.}
$$
Taking the <covariant divergence> of the <Einstein field equations>, with constant $G$, then gives $8\pi G\nabla^\mu T_{\mu\nu}=0$. The matter equation is therefore consistent with the geometric identity rather than imposing an extra incompatible condition on the metric.
<Stress-energy conservation> expresses the local balance of <energy> and <momentum>, including their transport and the effects of the gravitational <Levi-Civita connection>. In a freely falling frame at a point it reduces there to the familiar ordinary divergence law. A global conserved <energy> requires additional geometric structure, such as a timelike <Killing vector field>; it does not follow in an arbitrary <spacetime> just by replacing a <covariant derivative> with an ordinary derivative.
For <relativistic dust>,
$$
0=\nabla_\mu(\rho u^\mu u^\nu)
=u^\nu\nabla_\mu(\rho u^\mu)+\rho u^\mu\nabla_\mu u^\nu.
$$
With signature $(+,-,-,-)$, $u^\nu u_\nu=1$ implies $u_\nu u^\mu\nabla_\mu u^\nu=0$. Contracting gives $\nabla_\mu(\rho u^\mu)=0$; where $\rho>0$, substitution then gives $u^\mu\nabla_\mu u^\nu=0$. Thus the continuity law combines rest-<mass> conservation and free <geodesic> motion for this matter model.
For the <Newtonian limit>, take $c=1$ and $\eta=\operatorname{diag}(1,-1,-1,-1)$, consistently with Question 3. Use the static weak field sourced by the time-independent, slowly moving matter, retaining first order in $\epsilon$ and leading order in the spatial speed. Write spatial gradients with $\partial^i=\delta^{ij}\partial_j$ here; these are Euclidean spatial indices in the Newtonian equation. The <Levi-Civita connection> formula gives
$$
\Gamma_{00}{}^i=-\tfrac12\eta^{ij}\partial_j(\epsilon h_{00})
=\frac\epsilon2\delta^{ij}\partial_jh_{00}.
$$
The exact change from <proper time> to $t$ in the spatial <geodesic equation> is
$$
\frac{d^2x^i}{dt^2}
=-\Gamma_{\alpha\beta}{}^i v^\alpha v^\beta
+\Gamma_{\alpha\beta}{}^0v^\alpha v^\beta v^i,\qquad v^0=1.
$$
Terms containing spatial velocities are negligible at leading order, leaving
$$
\boxed{\frac{d^2x^i}{dt^2}=-\partial^i\Phi,\qquad
\Phi=\frac\epsilon2h_{00}.}
$$
An additive constant in $\Phi$ is immaterial to this acceleration and is normally fixed by the asymptotic time normalization. With units restored, $g_{00}=1+2\Phi_{\rm physical}/c^2$.
At the same order, staticity and the <curvature> definition give
$$
R_{00}=\partial_\lambda\Gamma_{00}{}^\lambda
-\partial_0\Gamma_{\lambda0}{}^\lambda+O(\epsilon^2)
=\frac\epsilon2\delta^{ij}\partial_i\partial_jh_{00}
=\nabla^2\Phi.
$$
The trace of the four-dimensional Einstein equations is $R=-8\pi GT$, so their trace-reversed form is
$$
R_{\mu\nu}=8\pi G\left(T_{\mu\nu}-\tfrac12g_{\mu\nu}T\right).
$$
For nonrelativistic dust, $u^0\simeq1$, $u^i\simeq0$, so $T_{00}\simeq\rho$ and $T\simeq\rho$. Keeping leading matter terms therefore gives $R_{00}=4\pi G\rho$. Combining this with the geometric expression yields
$$
\boxed{\nabla^2\Phi=4\pi G\rho.}
$$
This trace reversal is essential: equating $R_{00}$ directly to $8\pi G\rho$ would miss the factor of two. The approximation is the quasistatic Newtonian sector; arbitrary independently imposed gravitational radiation is not part of that sector.
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