= Solution
{c}
The <Ward identity> implies $q^\mu\Pi_{\mu\nu}=0$. <Lorentz invariance> then gives $\Pi_{\mu\nu}=(q^2g_{\mu\nu}-q_\mu q_\nu)\Pi$, so $g^{\mu\nu}\Pi_{\mu\nu}=(d-1)q^2\Pi$. The <gamma matrices> satisfy $\gamma_\mu\not k\gamma^\mu=(2-d)\not k$. Odd <traces> vanish, and therefore
$$
g^{\mu\nu}\operatorname{tr}[\gamma_\mu(\not k+m)\gamma_\nu(\not k+\not q+m)]
=4[-(d-2)k\cdot(k+q)+dm^2].
$$
Substituting gives the requested <scalar> coefficient:
$$
\boxed{\Pi(q^2)=\frac{4ie^2}{(d-1)q^2}\int_k
\frac{-(d-2)k\cdot(k+q)+dm^2}{[(k+q)^2-m^2+i0][k^2-m^2+i0]}.}
$$
To extract the <ultraviolet divergence> without introducing a <Feynman parameter>, set $A=(k+q)^2-m^2$, $B=k^2-m^2$ and use
$$
2k\cdot(k+q)=A+B+2m^2-q^2.
$$
Translation invariance of <dimensional regularization> gives $\int_kA^{-1}=\int_kB^{-1}=\mathcal I_1$. Consequently
$$
\Pi(q^2)=\frac{4ie^2}{(d-1)q^2}\left[-(d-2)\mathcal I_1+
\left(2m^2+\frac{d-2}{2}q^2\right)\mathcal I_2(q^2)\right].
$$
In the <pole> coefficient one may set $d=4$: the bracket becomes
$$
\frac{i}{16\pi^2\epsilon}[-2m^2+2m^2+q^2]
=\frac{iq^2}{16\pi^2\epsilon}.
$$
Thus the <ultraviolet pole of QED vacuum polarization> is
$$
\boxed{\Pi(q^2)\big|_{\rm UV}=-\frac{e^2}{12\pi^2\epsilon},\qquad
\Pi_{\mu\nu}\big|_{\rm UV}=-\frac{e^2}{12\pi^2\epsilon}(q^2g_{\mu\nu}-q_\mu q_\nu).}
$$
The cancellation of the apparent $m^2/q^2$ divergence is essential: <gauge invariance> forbids a <photon> <mass> <counterterm>. The scalar result at $q^2=0$ is understood by continuation after this cancellation.
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