= Solution
At $300\,\mathrm{MHz}$, $J/\mathrm{Hz}=300\,\Delta\delta/\mathrm{ppm}$. The six resolved <scalar spin-spin couplings> are
$$
\begin{aligned}
{}^3J_{bc}&=(5.893-5.842)300=\boxed{15.3\,\mathrm{Hz}},\\
{}^3J_{ab}&=(7.071-7.048)300=\boxed{6.9\,\mathrm{Hz}},\\
{}^4J_{ac}&=(5.848-5.842)300=\boxed{1.8\,\mathrm{Hz}},\\
{}^3J_{de}&=(4.932-4.885)300=\boxed{14.1\,\mathrm{Hz}},\\
{}^3J_{df}&=(4.589-4.568)300=\boxed{6.3\,\mathrm{Hz}},\\
{}^2J_{ef}&=(4.932-4.927)300=\boxed{1.5\,\mathrm{Hz}}.
\end{aligned}
$$
For example, the two quartets of $c$ are $\{5.842,5.848,5.854,5.860\}$ and $\{5.893,5.899,5.905,5.911\}$ ppm. Their origins differ by $0.051$ ppm, while the spacing within either quartet is $0.006$ ppm. This separates the trans alkene interaction from the smaller allylic methyl interaction.
The four $d$ lines independently give $(7.377-7.330)300=14.1$ and $(7.377-7.356)300=6.3\,\mathrm{Hz}$. The large interaction identifies $e$ as trans to $d$; $f$ has the smaller cis interaction. The double-headed arrows in the structural map identify the actual interacting nuclei. Coupling to $a$ means the same coupling to each of its three equivalent hydrogens.
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