Stability of branched alkenes
The Z isomer is destabilized by comparison to the E isomer due to steric strain. This is a consequence of the isopropyl groups being on the same side of the double bond in the Z isomer, resulting in electron-electron repulsion between the groups. This also, at least partially, accounts for the increase in the bond angle that should be observed in the Z isomer: increasing that angle also increases the distance between the isopropyl groups, minimizing the mutual repulsion between them (admittedly, this is at the expense of the other angles, but the bulkiness of the isopropyls by comparison to hydrogens necessitates that distortion from idealized geometry). A similar effect occurs with the geminal isopropyl groups in the 1,1- isomer.

Additionally, the degree of substitution of an alkene has a dramatic effect on stability. The presence of alkyl groups on the carbons of an alkene is generally stabilizing (as can be empirically confirmed by examining thermodynamic data, i.e., heats of hydrogenation and heats of combustion). Evidently, calculations grounded in molecular orbital theory indicate that the mixing of $\sigma$ type molecular orbitals of alkanes with the $\pi$ orbitals of less-substituted alkenes is typically energy-lowering overall, and that the $\pi^{*}$ antibonding orbitals are sufficiently close in energy to the $\sigma$ orbitals of alkanes that hyperconjugative stabilization occurs (albeit relatively small in magnitude, since the energies are not very close). In the 1,1- isomer, only one carbon of the alkene can be stabilized by hyperconjugation, while both carbons should be able to benefit from it in the 1,2- isomers. (Speculatively, I would also think that the extent of orbital overlap should be poorer in the 1,1- isomer, since the geminal isopropyl groups would likely be staggered with respect to one another, which might prevent proper alignment of the alkyl group orbitals with those of the alkene $\pi$ system.)