= Solution
Use the <NMR coupling constant> conversion $J/\mathrm{Hz}=(\Delta\delta/\mathrm{ppm})(\nu_0/\mathrm{MHz})$. At $300\,\mathrm{MHz}$, separating the overlapping quartets gives
$$
\begin{aligned}
{}^3J_{\mathrm H_a\mathrm H_b}&=(6.780-6.728)\,300=\boxed{15.6\,\mathrm{Hz}},\\
{}^3J_{\mathrm H_a\mathrm F}&=(6.750-6.728)\,300=\boxed{6.6\,\mathrm{Hz}},\\
{}^4J_{\mathrm H_b\mathrm F}&=(6.467-6.461)\,300=\boxed{1.8\,\mathrm{Hz}},\\
{}^3J_{\mathrm H_c\mathrm H_d}&=(1.355-1.331)\,300=\boxed{7.2\,\mathrm{Hz}}.
\end{aligned}
$$
The $\mathrm H_b$ doublet separation independently gives $(6.513-6.461)300=15.6\,\mathrm{Hz}$. For $\mathrm H_a$, the two quartets are $\{6.728,6.750,6.772,6.794\}$ and $\{6.780,6.802,6.824,6.846\}$ ppm; simply taking adjacent lines in their combined ordering would mix different <scalar spin-spin couplings>.
The methylene quartet gives an average adjacent spacing $(4.317-4.244)/3=0.02433$ ppm, or $7.3\,\mathrm{Hz}$. Thus its printed rounded positions agree with the methyl-derived value to about $0.1\,\mathrm{Hz}$; quote the ethyl <NMR coupling constant> as approximately $7.2$–$7.3\,\mathrm{Hz}$. The large vinylic value supports the trans arrangement. The arrows below join the interacting nuclei; each arrow to $\mathrm F_3$ represents the same interaction with each equivalent <fluorine> nucleus.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025-nmr-couplings.png]
{title=Calculated <scalar spin-spin couplings> for the vinylic, ethyl and trifluoromethyl sites}
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