Well, OP is asking a rather demanding question. I can't answer all of it, but I can address the observed-physics part of it.
Nuclear decay can be influenced by the atomic electron shell in the case of electron capture. At the nucleon level the reaction is $\mathrm{p^+}+\mathrm{e}^- \longrightarrow \mathrm{n^0} + \nu_\mathrm{e}$. As you might expect, the decay rate is proportional to the product of the nucleon (ie proton) density and the electron density. The nucleus is typically 1000 times smaller in diameter than the electron shell. For electrons in the K-shell (s-orbital), the density is maximal at r=0 (the locus of the nucleus), and it decreases away from the nucleus. For orbitals with higher momentum, the electron density at the nucleus is essentially zero, being proportional to $r^n$, where $n=1,2,3,...$ for p,d,f, and higher orbitals.
So, electron capture in the nucleus can be influenced by a chemical bonding (compared to the lone atom) if that chemical bonding changes the s-orbital. In practice this occurs only in the lightest elements. In the heavier elements the s-orbitals are shielded by too many layers of other electrons.
There are a few unusual cases. $^{163}_{66}\mathrm{Dy}$ is a stable atom. But when fully ionized, the bare nucleus can $\beta$-decay into a bound state where the electron appears in the 1s orbital (K-shell).
Similarly for $^{187}_{75}\mathrm{Re}$. As an atom it has a decay half life of $42\cdot 10^9$ years. When fully ionized, its half life shortens dramatically to 47 days due to bound-state $\beta^−$ decay into the K and L shells.
Finally, $^7_4Be$ decays by electron capture only. If you use a laser to excite its outer electrons, the s-orbital will be affected a little ($<1\%$) and the half life will be changed accordingly, depending on what fraction of the atoms is excited or ionized.
Sources: Beta decay, Changing decay rates