= Solution
Each formula unit of $\mathrm{CF}_{0.8}$ accepts $0.8$ electrons. Using the supplied Faraday constant and the periodic-table masses,
$$
M=12.011+0.8(18.998)=27.2094\,\mathrm{g\,mol^{-1}},
$$
$$
\boxed{q_{\rm active}=\frac{0.8F}{3.6M}
\simeq788.5\,\mathrm{mAh\,g^{-1}}}.
$$
The factor $3.6$ converts coulombs to mAh. This value is per gram of the \b[active fluorinated carbon], not per gram of the entire electrode.
\b[The composite value is underdetermined: the mass of active CF₀.₈ is not supplied.] If that mass is $m$ grams, and the carbon black and binder are inactive in the stated conversion reaction, then
$$
\boxed{q_{\rm composite}=788.5\frac{m}{m+2.0+0.2}
=788.5\frac{m}{m+2.2}\,\mathrm{mAh\,g^{-1}}}.
$$
This uses cathode composite mass and excludes the lithium counterelectrode and hardware. No unique numerical composite capacity follows from the two additive masses alone; assuming an unprovided active mass would change the requested quantity.
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