= Solution
For adjoint-valued quantities write $(X\times Y)_a=\varepsilon_{abc}X_bY_c$. Substituting the infinitesimal transformation into the <gauge field strength> and using the <Jacobi identity> gives
$$
\boxed{\delta F_{\mu\nu a}=g\varepsilon_{abc}F_{\mu\nu b}\lambda_c.}
$$
A useful way to organize the calculation is $[D_\mu,D_\nu]X=gF_{\mu\nu}\times X$: the <gauge field strength> transforms covariantly because the <gauge covariant derivative> does. Thus $\delta(F_a^{\mu\nu}F_{\mu\nu a})=2gF_a^{\mu\nu}\varepsilon_{abc}F_{\mu\nu b}\lambda_c=0$ by antisymmetry. This proves \b[<gauge invariance> of the <Yang-Mills action>].
The <Yang-Mills equations> are $D_\mu F^{\mu\nu}=0$. They determine physical evolution only modulo <gauge redundancy>: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In <Hamiltonian mechanics>, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a <Gauss law constraint in gauge theory>. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later <potential energy>, so the <potential energy> itself is not uniquely fixed by initial physical data.
In the uncorrected <path integral>, integration along <gauge orbits> overcounts equivalent fields. At the perturbative level the quadratic operator of the <kinetic term> has gauge zero modes and no inverse, so it does not supply a <quantum field theory propagator>. <Gauge fixing> together with the <Faddeev-Popov determinant> removes this obstruction in the local perturbative construction.
For the odd <BRST transformation>, products obey the <graded Leibniz rule>. The adjoint product with <Grassmann variables> satisfies
$$
X\times Y=-(-1)^{|X||Y|}Y\times X.
$$
In particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations give
$$
s^2A_\mu=D_\mu(sc)+g(sA_\mu)\times c
=-\frac g2D_\mu(c\times c)+g(D_\mu c)\times c=0,
$$
since the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,
$$
s^2c=-\frac g2\bigl((sc)\times c-c\times(sc)\bigr)
=-g(sc)\times c=\frac{g^2}{2}(c\times c)\times c=0.
$$
The final equality is the <graded Jacobi identity>; in components it is the <Jacobi identity> for $\varepsilon_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar c=-sb=0$ and $s^2b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence \b[$s^2=0$ on every polynomial in the fields], off shell, without imposing field equations.
Let $\Psi=\int d^dx\,[\partial^\mu\bar c_aA_{\mu a}+\xi\bar c_ab_a/2]$ be the <gauge-fixing fermion>. Since $sF=gF\times c$, $s\mathcal L_{\mathrm{YM}}=0$. The <BRST-exact operator> in $\mathcal L_q=\mathcal L_{\mathrm{YM}}-s\psi$ has zero variation by <BRST nilpotence>, so
$$
\boxed{s\mathcal L_q=0.}
$$
To see the resulting kinetic terms and signs, expand the exact term with $s\bar c=-b$:
$$
-s\psi=(\partial^\mu b_a)A_{\mu a}
+(\partial^\mu\bar c_a)(D_\mu c)_a+\frac\xi2 b_a^2.
$$
After <integration by parts>,
$$
\mathcal L_q=\mathcal L_{\mathrm{YM}}-b_a\partial^\mu A_{\mu a}
+\frac\xi2b_a^2-\bar c_a\partial^\mu(D_\mu c)_a.
$$
The auxiliary <Nakanishi-Lautrup field> $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial\cdot A_a)^2/(2\xi)$ for $\xi\ne0$. At $\xi=0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The <Grassmann Gaussian integral> gives the <determinant> of the gauge-condition operator $-\partial\cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.
At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic <kinetic term>. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using <Wick contractions> gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed <action> $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.
At $\xi=1$, the free gauge <action>, up to a boundary term, is $\tfrac12\int A_{\mu a}\Box A_a^\mu$. Its Fourier kernel is $-p^2\eta_{\mu\nu}\delta_{ab}$. With the mostly-plus convention used above, the <Feynman-gauge adjoint propagator> is
$$
\boxed{\langle T A_{\mu a}(x)A_{\nu b}(y)\rangle_0
=\int\frac{d^dp}{(2\pi)^d}\,e^{ip\cdot(x-y)}
\frac{-i\eta_{\mu\nu}\delta_{ab}}{p^2-i\epsilon}.}
$$
This is consistent with the scalar-line convention in Question 2.
Finally define the odd functional $B=\tfrac12\int d^dx\,\bar c_ab_a$. Then $\partial_\xi S_q=-sB$. Differentiating the normalized expectation, including its denominator, gives
$$
\partial_\xi\langle X\rangle
=-i\left[\langle XsB\rangle-\langle X\rangle\langle sB\rangle\right]
$$
for an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is <BRST-closed>, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed <BRST Ward identity> gives $\langle sB\rangle=0$ and $\langle s(XB)\rangle=0$. For an even $X$ with $sX=0$, the latter says $\langle XsB\rangle=0$. Therefore
$$
\boxed{\partial_\xi\langle X\rangle=0.}
$$
The same conclusion holds for any <BRST-closed> insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward identity.
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