= Solution
The basic <Feynman diagrams> have $\tau^-\to\nu_\tau W^{-*}$ followed by $W^{-*}\to e^-\bar\nu_e$, $\mu^-\bar\nu_\mu$, or $d\bar u,s\bar u$, with the latter pairs undergoing <hadronization>. See the charged-current diagrams in <2/a>. Neglect all daughter <masses>, <QCD> corrections and inaccessible charm final states, and assume <lepton universality>. Each leptonic mode then has one unit of <decay width>; the two accessible <quark> combinations have three <color charge> states and total CKM weight $|V_{ud}|^2+|V_{us}|^2\simeq1$. Therefore
$$
\Gamma_e:\Gamma_\mu:\Gamma_{\rm had}=1:1:3,\qquad B_{\rm had}=\frac35=0.60.
$$
The result is an inclusive <parton model> estimate. Perturbative <QCD> corrections increase the hadronic contribution, while <muon mass>, nonperturbative dynamics and radiative corrections also matter; the physical fraction is roughly $0.65$.
Every ordinary hadronic mode already contains a <tau neutrino>, so it contains a <lepton> as well as <hadrons> if invisible particles are counted. For an additional visible charged <lepton>, a virtual <photon> can internally convert in a hadronic mode, for example $\tau^-\to\nu_\tau\pi^-\gamma^*\to\nu_\tau\pi^-e^+e^-$. The extra electromagnetic vertices suppress such modes; they are not a large fraction of the inclusive width. Alternatively a daughter <pion> or <kaon> can decay semileptonically or leptonically in flight, for example $\tau^-\to\nu_\tau\pi^-K^0$ followed by a semileptonic neutral-<kaon> decay. Those are secondary <decay chains>, not independent primary <tau lepton> branching fractions. Lepton-flavour-violating neutrinoless modes such as $\tau^-\to\mu^-\pi^+\pi^-$ do not contribute appreciably in the <Standard Model>.
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