= Solution
The two outward surface normals point in opposite directions. With positive <heat flux> defined as heat leaving each face, <Fourier's law> gives
$$
q''_{\rm left}=kT'(0),\qquad q''_{\rm right}=-kT'(a).
$$
Using the derivative from part(a), \b[the removed fluxes per unit face area] are
$$
\boxed{q''_{\rm left}=\frac{q_0}{\mu}\left[1-\frac{1-e^{-\mu a}}{\mu a}\right]
=\frac{q_0}{\mu}(1-C),}
$$
$$
\boxed{q''_{\rm right}=\frac{q_0}{\mu}\left[\frac{1-e^{-\mu a}}{\mu a}-e^{-\mu a}\right]
=\frac{q_0}{\mu}(C-e^{-\beta}).}
$$
Both are positive because $e^{-\beta}<C<1$. Their sum satisfies the independent <conservation of energy> check
$$
q''_{\rm left}+q''_{\rm right}
=\frac{q_0}{\mu}(1-e^{-\mu a})
=\int_0^a q_0e^{-\mu x}\,dx.
$$
The illuminated face removes more <heat> because heating is concentrated nearer it. Equivalently, weighting each source slice by the fraction of its heat conducted to either equal-temperature boundary gives
$$
q''_{\rm left}=\int_0^a q'''(x)\frac{a-x}{a}\,dx,
\qquad
q''_{\rm right}=\int_0^a q'''(x)\frac{x}{a}\,dx.
$$
A decreasing source weights the left face more strongly. In the uniform-heating limit, \b[each face removes $q_0a/2$]. If a heat removal rate rather than a flux is wanted, multiply the relevant expression by the face area.
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