= Solution
The <Adjoint representation of a Lie algebra> acts on the algebra itself: $\operatorname{ad}_X(Y)=[X,Y]$. In the basis $T_b$, its <matrices> are
$$
\boxed{(T_a^{\rm ad})^c{}_b=c^c{}_{ab}.}
$$
For any algebra element $X$, the <Jacobi identity> gives
$$
[T_a^{\rm ad},T_b^{\rm ad}]X=[T_a,[T_b,X]]-[T_b,[T_a,X]]=[[T_a,T_b],X].
$$
Therefore the $D\times D$ <matrices> satisfy the same <Lie bracket> relations:
$$
\boxed{[T_a^{\rm ad},T_b^{\rm ad}]=c^c{}_{ab}T_c^{\rm ad}.}
$$
This construction does not require that the algebra be simple or that the adjoint action be faithful.
For the field calculation, write $A_\mu=A_\mu^aT_a$ and $\lambda=\lambda^aT_a$. In the derivative-plus-connection convention the <Yang-Mills field strength> is
$$
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu+[A_\mu,A_\nu],\qquad
\delta A_\mu=\partial_\mu\lambda+[A_\mu,\lambda].
$$
The sign of the parameter here is the negative of the matter-field parameter in the usual <gauge transformation in the derivative-plus-connection convention>. Work with the stated $\lambda$ throughout. Expanding the variation gives
$$
\begin{aligned}
\delta F_{\mu\nu}={}&[\partial_\mu A_\nu-\partial_\nu A_\mu,\lambda]
+[A_\nu,\partial_\mu\lambda]-[A_\mu,\partial_\nu\lambda]\\
&+[\partial_\mu\lambda,A_\nu]+[A_\mu,\partial_\nu\lambda]
+[[A_\mu,\lambda],A_\nu]+[A_\mu,[A_\nu,\lambda]].
\end{aligned}
$$
The second derivatives cancel by commutativity of partial derivatives, and all remaining derivatives of $\lambda$ cancel by antisymmetry of the <Lie bracket>. The <Jacobi identity> makes the last two terms $[[A_\mu,A_\nu],\lambda]$. Hence
$$
\boxed{\delta F_{\mu\nu}=[F_{\mu\nu},\lambda],\qquad
\delta F^a_{\mu\nu}=c^a{}_{bc}F^b_{\mu\nu}\lambda^c.}
$$
Unlike the connection, the curvature has no inhomogeneous derivative term: it transforms in the <Adjoint representation>.
The quadratic form $\kappa_{ab}=\operatorname{tr}(T_a^{\rm ad}T_b^{\rm ad})$ is the <Killing form>. Its symmetry follows from cyclicity of the <matrix trace>. Using the just-proved adjoint commutators,
$$
\begin{aligned}
c^d{}_{ca}\kappa_{db}+c^d{}_{cb}\kappa_{ad}
&=\operatorname{tr}\left([T_c^{\rm ad},T_a^{\rm ad}]T_b^{\rm ad}
+T_a^{\rm ad}[T_c^{\rm ad},T_b^{\rm ad}]\right)\\
&=\operatorname{tr}[T_c^{\rm ad},T_a^{\rm ad}T_b^{\rm ad}]=0.
\end{aligned}
$$
Thus
$$
\boxed{c^d{}_{ca}\kappa_{db}+c^d{}_{cb}\kappa_{ad}=0.}
$$
Equivalently, each adjoint generator is skew with respect to this invariant bilinear form. This invariance holds even when the <Killing form> is degenerate; no inverse form is needed here.
The spacetime metric used to raise Lorentz indices is unaffected by the internal <gauge transformation>. Since $\delta F=-\operatorname{ad}_\lambda F$, invariance gives
$$
\begin{aligned}
\delta\bigl(\kappa_{ab}F^{a\mu\nu}F^b_{\mu\nu}\bigr)
&=\kappa(\delta F^{\mu\nu},F_{\mu\nu})+\kappa(F^{\mu\nu},\delta F_{\mu\nu})\\
&=-\lambda^c\left[c^d{}_{ca}\kappa_{db}+c^d{}_{cb}\kappa_{ad}\right]
F^{a\mu\nu}F^b_{\mu\nu}=0.
\end{aligned}
$$
Therefore \b[the Killing-form contraction of the field strengths is gauge invariant], as required for the internal contraction in a <Yang-Mills action>.
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