Varying the <Ricci tensor> and contracting gives
$$
\delta R=-R^{ab}h_{ab}
+\nabla_a\!\left(g^{cb}\delta\Gamma^a{}_{cb}
-g^{ab}\delta\Gamma^c{}_{cb}\right).
$$
Substitution of part (i), followed by <metric compatibility>, reduces the divergence to
$$
\nabla_a\nabla_bh^{ab}-\nabla^a\nabla_ah.
$$
Thus the <metric variation of scalar curvature> is
$$
\boxed{\delta R=-R^{ab}\delta g_{ab}
-g^{ab}\nabla_c\nabla^c\delta g_{ab}
+\nabla^a\nabla^b\delta g_{ab}},
$$
and consequently
$$
\boxed{\alpha=-1,
\qquad \beta=1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 309 / 3 / a / ii / Solution

For $h_{ab}=\delta g_{ab}$, the <Ricci scalar> varies as
$$
\delta R=-R^{ab}h_{ab}+\nabla_a\nabla_bh^{ab}-\nabla^a\nabla_ah,
\qquad h=g^{ab}h_{ab}.
$$
The last two terms form a <covariant divergence> and hence contribute only a boundary term to a compactly supported variation of the <Einstein-Hilbert action>.

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 Incoming links: Metric variation of scalar curvature