= Solution
<Lorentz covariance>[Lorentz-covariant] <Lorentz-covariant rank-two loop reduction>[tensor integration] replaces $q_\mu q_\nu$ by $g_{\mu\nu}q^2/d$. The numerator consequently contributes $(d-1)k^2q^2/d$. The supplied master integral with powers $a=1,b=3$ gives
$$
\int\frac{d^dq}{(2\pi)^d}\frac{q^2}{(q^2+D)^3}
=\frac{i}{(4\pi)^{d/2}}\frac d4\Gamma(\epsilon/2)D^{-\epsilon/2}.
$$
Inserting the corrected parameter integral gives
$$
i\Pi_{\mathrm{ex}}=\frac{2i(d-1)e^2\mu^\epsilon k^2}{(4\pi)^{d/2}}\Gamma(\epsilon/2)\int_0^1(1-x)D^{-\epsilon/2}dx.
$$
The pole uses $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, $D^{-\epsilon/2}=1+O(\epsilon)$ and $\int_0^1(1-x)dx=1/2$. Thus
$$
\boxed{i\Pi_{\mathrm{loop,div}}=i\frac{3e^2}{8\pi^2\epsilon}k^2.}
$$
There is no momentum-independent pole at $\lambda=0$ in this transverse gauge after removing the <Tadpole diagram>[photon tadpole]. The <Two-point function>[two-point] <counterterm> insertion is $-i(\delta Z_\phi k^2+\delta M^2)$, so <Minimal subtraction scheme>[minimal subtraction] requires
$$
\boxed{\delta Z_\phi=\frac{3e^2}{8\pi^2\epsilon},\qquad \delta M^2=0\qquad(d=4-\epsilon).}
$$
These coefficients cancel the actual <Loop diagram>[one-loop] insertion derived from the <action>. Applying the misprinted weight in part e would give a wave-function pole four times too small. In a $d=4-2\varepsilon$ convention the same answer is $\delta Z_\phi=3e^2/(16\pi^2\varepsilon)$. Finally, $\delta M^2=0$ does not mean that the bare <Mass>[mass parameter] is unchanged: $Z_\phi m_0^2=m^2$ gives $m_0^2-m^2=-m^2\delta Z_\phi+O(e^4)$. <Photon>, charge and <Quartic interaction>[quartic] <counterterms> require their respective <Green functions> and cannot be determined from this <Scalar field>[scalar] <two-point function> alone.
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