= Solution
For the torsion-free <Levi-Civita connection>, cyclically summing the displayed coordinate expression and using $\Gamma^a{}_{bc}=\Gamma^a{}_{cb}$ gives the algebraic <first Bianchi identity>
$$
\boxed{R^a{}_{bcd}+R^a{}_{cdb}+R^a{}_{dbc}=0}.
$$
Apply the <Jacobi identity> to three covariant-derivative commutators, or differentiate the coordinate formula in <normal coordinates>. The third-derivative terms and the derivatives of quadratic connection terms cancel cyclically, giving the differential <second Bianchi identity>
$$
\boxed{\nabla_eR^a{}_{bcd}
+\nabla_cR^a{}_{bde}
+\nabla_dR^a{}_{bec}=0}.
$$
Contracting the first and third curvature indices and then contracting once more yields
$$
\nabla^aR_{ab}-\frac12\nabla_bR=0,
$$
or the <contracted Bianchi identity>
$$
\boxed{\nabla^aG_{ab}=0}.
$$
For the metric variation $h_{ab}=\delta g_{ab}$,
$$
\delta\sqrt{-g}=\frac12\sqrt{-g}\,h,
$$
while the two double-divergence terms in the supplied <metric variation of scalar curvature> become a boundary term after <integration by parts for tensor fields>. The <Einstein-Hilbert action> therefore contributes $G_{ab}+\Lambda g_{ab}$. Varying the <Proca action> with respect to the metric gives the <Proca stress-energy tensor>, and the full <Einstein-Proca theory> equation is
$$
\boxed{
R_{ab}-\frac12Rg_{ab}+\Lambda g_{ab}
=8\pi\left[
F_{ac}F_b{}^c-\frac14g_{ab}F_{cd}F^{cd}
+m^2A_aA_b-\frac12m^2g_{ab}A_cA^c
\right]}.
$$
Since $\delta F_{ab}=2\nabla_{[a}\delta A_{b]}$, variation with respect to $A_b$ and one <integration by parts> give the covariant <Proca equation>
$$
\boxed{\nabla_aF^{ab}-m^2A^b=0}.
$$
Taking its <covariant divergence> gives
$$
m^2\nabla_bA^b=\nabla_b\nabla_aF^{ab}
=\frac12[\nabla_b,\nabla_a]F^{ab}=0,
$$
because the resulting contractions pair the symmetric <Ricci tensor> with the antisymmetric <electromagnetic field tensor>. Since $m\ne0$, the <Lorenz constraint in Proca theory> follows:
$$
\boxed{\nabla_aA^a=0}.
$$
Moreover, $F=dA$ and $d^2=0$, so the <Electromagnetic Bianchi identity> is
$$
\boxed{\nabla_aF_{bc}+\nabla_bF_{ca}+\nabla_cF_{ab}=0}.
$$
Finally, use that identity to differentiate the matter tensor. The result groups naturally as
$$
\nabla^aT_{ab}
=F_b{}^c\left(\nabla^aF_{ac}-m^2A_c\right)
+m^2A_b\nabla_aA^a.
$$
Both terms vanish by the <Proca equation> and its <Lorenz constraint in Proca theory>. Thus $\nabla^aT_{ab}=0$, exactly as required by the <contracted Bianchi identity> applied to the <Einstein field equations>. The vector equation is therefore consistent with the gravitational equation.
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