= Solution
\b[Flavour issue in the printed effective interaction.] The displayed charged-current factor uses an <Electron> field even for the muon-<neutrino> term. Taken literally, this introduces an extra $\nu_\mu e$ charged-current interaction, contradicting the Standard Model diagrams in part (a). In the Standard Model that factor uses the charged lepton of the same flavour; for <Electron> scattering the charged-current contribution is therefore present only for $\nu_e$. The <weak flavour selection in neutrino-electron scattering> fixes the physically consistent assignment used below. The literal printed consequence is given afterwards, so the discrepancy is explicit.
At leading order in the <Electron> mass and in the four-fermion regime, define $\Gamma_L^\alpha=\gamma^\alpha(1-\gamma^5)$. The <weak neutral current> yields the <scattering amplitude>
$$
\boxed{\mathcal M_\mu=-\frac{G_F}{\sqrt2}
[\bar u_\nu(k')\Gamma_L^\alpha u_\nu(k)]
[\bar u_e(p')\gamma_\alpha(c_V-c_A\gamma^5)u_e(p)].}
$$
An overall amplitude phase has no observable effect. The <Fermi interaction> is valid when the exchanged invariants are small compared with the squared weak-boson masses.
Average over the two initial <Electron> spin states, and sum over final spins. There is no averaging over a sterile right-handed incoming <neutrino>. Using the <fermion spin sum>, the averaged square is
$$
\overline{|\mathcal M_\mu|^2}=\frac{G_F^2}{4}
\operatorname{Tr}[\not k'\Gamma_L^\alpha\not k\Gamma_L^\beta]\,
\operatorname{Tr}[\not p'\gamma_\alpha(c_V-c_A\gamma^5)
\not p\gamma_\beta(c_V-c_A\gamma^5)].
$$
To display the trace contraction, define
$$
S^{\alpha\beta}(a,b)=a^\alpha b^\beta+a^\beta b^\alpha-g^{\alpha\beta}(a\cdot b),
\qquad E^{\alpha\beta}(a,b)=\epsilon^{\alpha\beta\rho\sigma}a_\rho b_\sigma.
$$
The supplied <gamma matrix trace identities> give the two tensors $8(S(k',k)+iE(k',k))$ and $4[(c_V^2+c_A^2)S(p',p)+2ic_Vc_AE(p',p)]$. Symmetric-antisymmetric mixed contractions vanish. Set $X=(k'\cdot p')(k\cdot p)$ and $Y=(k'\cdot p)(k\cdot p')$. The remaining contractions are $S(k',k)S(p',p)=2(X+Y)$ and $E(k',k)E(p',p)=-2(X-Y)$, using the stated negative epsilon-contraction sign. Their sum gives
$$
\overline{|\mathcal M_\mu|^2}
=16G_F^2\left[
(c_V+c_A)^2(k\cdot p)(k'\cdot p')
+(c_V-c_A)^2(k\cdot p')(k'\cdot p)\right].
$$
The massless <Mandelstam variables> satisfy $s+t+u=0$, and the two products are $s^2/4$ and $u^2/4$, respectively. Therefore
$$
\boxed{\overline{|\mathcal M_\mu|^2}
=4G_F^2\bigl[(c_V+c_A)^2s^2+(c_V-c_A)^2u^2\bigr].}
$$
In the centre-of-momentum frame, momentum conservation gives $E=\sqrt s/2$. If $z=\cos\theta$ is the angle cosine between $p$ and $p'$, then $u=-s(1+z)/2$. The massless <two-body Lorentz-invariant phase space> is $d\rho_2=d\Omega/(32\pi^2)$, while the flux is $2s$. Thus
$$
\frac{d\sigma}{dz}=\frac{\overline{|\mathcal M|^2}}{32\pi s}
=\frac{G_F^2s}{8\pi}
\left[(c_V+c_A)^2+\frac14(c_V-c_A)^2(1+z)^2\right].
$$
Integrate $\int_{-1}^1dz=2$ and $\int_{-1}^1(1+z)^2dz=8/3$:
$$
\boxed{\sigma_\mu=\frac{G_F^2s}{4\pi}
\left[(c_V+c_A)^2+\frac13(c_V-c_A)^2\right]
=\frac{G_F^2s}{3\pi}(c_V^2+c_Vc_A+c_A^2).}
$$
Consequently the intended coefficients are
$$
\boxed{F(s)=\frac{s}{3\pi},\qquad A=1,\qquad B=1,\qquad C=0.}
$$
This is the <massless neutrino-electron contact cross-section>.
For completeness, the literal printed muon-<neutrino> charged-current term would instead shift both couplings by one, exactly as in part (c):
$$
\sigma_{\mu,\mathrm{literal}}
=\frac{G_F^2s}{3\pi}
[c_V^2+c_Vc_A+c_A^2+3(c_V+c_A)+3].
$$
For arbitrary real $c_V,c_A$, its linear terms cannot be represented by the requested quadratic form with coupling-independent constants. \b[The Standard Model result and the literal printed interaction are not the same problem.]
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