= Solution
Choose the convention $\psi'=g\psi$ for a field in the defining representation and $D_\mu=\partial_\mu+A_\mu$. Requiring $D_\mu'\psi'=gD_\mu\psi$ gives the <Yang-Mills gauge transformation>
$$
\boxed{A_\mu'=gA_\mu g^{-1}-(\partial_\mu g)g^{-1}.}
$$
The <gauge field strength> is the curvature $[D_\mu,D_\nu]=F_{\mu\nu}$, so
$$
\boxed{F_{\mu\nu}'=gF_{\mu\nu}g^{-1}.}
$$
For an adjoint field $Y$, the <adjoint covariant derivative> is $D_\mu Y=\partial_\mu Y+[A_\mu,Y]$, and direct substitution gives $D_\mu'Y'=g(D_\mu Y)g^{-1}$. Consequently $D_\mu F_{\nu\rho}$ transforms by conjugation too.
The <Killing-form invariance under a Lie-group adjoint action> gives $\kappa(gYg^{-1},gZg^{-1})=\kappa(Y,Z)$. Its proof is that the adjoint matrices are conjugated by the same invertible map, leaving their trace product unchanged. With $s=\kappa(F_{\mu\nu},F^{\mu\nu})$, both $s^p$ and $\kappa(D_\mu F_{\nu\rho},D^\mu F^{\nu\rho})$ are therefore gauge invariant.
For their field equations, take a compactly supported variation $a_\mu=\delta A_\mu$. Curvature varies by
$$
\delta F_{\mu\nu}=D_\mu a_\nu-D_\nu a_\mu.
$$
The invariant form permits <gauge-covariant integration by parts>, because $\partial_\mu\kappa(Y,Z)=\kappa(D_\mu Y,Z)+\kappa(Y,D_\mu Z)$. Antisymmetry of $F$ gives
$$
\delta(s^p)=4p\,s^{p-1}\kappa(D_\mu a_\nu,F^{\mu\nu}).
$$
Integrating, the <power of the Yang-Mills invariant> gives the equation
$$
\boxed{D_\mu\bigl(s^{p-1}F^{\mu\nu}\bigr)=0.}
$$
For $p=1$ this recovers the <Yang-Mills equations>. For larger $p$, keep the displayed product form, including at $s=0$.
For the derivative-squared action, write $B_{\mu\nu\rho}=D_\mu F_{\nu\rho}$. Varying the derivative as well as the curvature gives
$$
\delta B_{\mu\nu\rho}=D_\mu\delta F_{\nu\rho}+[a_\mu,F_{\nu\rho}].
$$
Its first variation is
$$
\delta S_2=2\int\kappa(D^\mu F^{\nu\rho},D_\mu\delta F_{\nu\rho})\,d^4x
+2\int\kappa(D^\mu F^{\nu\rho},[a_\mu,F_{\nu\rho}])\,d^4x.
$$
Integrate the first term once, substitute $\delta F_{\nu\rho}=D_\nu a_\rho-D_\rho a_\nu$, and integrate again. It becomes $4\int\kappa(a_\nu,D_\mu D^2F^{\mu\nu})d^4x$, where $D^2=D_\lambda D^\lambda$. Invariance converts the second term into $2\int\kappa(a_\nu,[F_{\rho\sigma},D^\nu F^{\rho\sigma}])d^4x$. Nondegeneracy of the Killing form and arbitrary variations yield the <derivative-squared Yang-Mills action> equation
$$
\boxed{2D_\mu D_\lambda D^\lambda F^{\mu\nu}+[F_{\rho\sigma},D^\nu F^{\rho\sigma}]=0.}
$$
The derivative order matters because covariant derivatives generally do not commute. In an abelian limit the commutator term vanishes and the equation reduces to $\Box\partial_\mu F^{\mu\nu}=0$, a useful check on the fourth-order equation for the potential.
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