Why does mesitylene absorb at a longer wavelength than benzene (UV-Vis)?
The general observation is that alkyl substitution intensifies and shifts the benzene spectrum to longer wavelengths. The effect is similar to that of groups with lone pairs such as $\ce{NH2}$ and O-alkyl. So I don't think that opposite effects are observed as you suggest. Some data are given below.

The 0-0 band transition, from  v=0 vibrational level to v' =0 in the first excited state of benzene is at 262 nm (due to symmetry this is a forbidden transition so its position has to be calculated. The maximum in the S0-S1 transition in benzene spectrum is at 245nm, and S0-S2 at 203 nm).  Toluene does not have a forbidden 0-0 transition, and has a 0-0 band at approx 266 nm (with maxima at 261 nm and S0-S2 at 207 nm), 0-zylene 0-0 is at approx 273 nm, so the substituents do stabilise the excited state very slightly. This has been attributed to 'hyperconjugation', i.e. some charge being donated by the methyl group to the ring, or the alkyl groups behaves as it it had a lone pair. (The idea of hyperconjugation is still hotly contested). Toluene shows a slight increase in intensity as the symmetry is broken $\ce{D_{6h} to C_{2v}}$ but otherwise the effect on the electronic spectrum is slight. Aniline has a relatively broad and intense  absorption maximum (not 0-0) at 280 nm, nitrobenzene max very broad spectrum with a maximum at approx 268nm, but with a long shoulder in absorption stretching out to almost 400 nm, which is thought to cover the n-$\pi^*$ transition.

With very broad and structureless spectra (almost all the molecules you mention except benzene and toluene) the maximum intensity does usually always correspond to the 0-0 energy of the transition, due to Franck-Condon factors (overlap of wavefunctions which control intensity of transitions) and the fact that extra groups add considerable vibrational intensity to transitions and so broaden them. Thus comparing absorption peaks is not a reliable way of comparing state energies. (Experimentally the only ways are to measure vapour  phase absorption spectra or measure in a solid matrix  at 4k if possible.) Secondly spectra are stabilised differently in different solvents, even relatively non-polar ones and this can lead to spectral shifts.  

The general sort of handwaving explanation, that is nevertheless useful,  is to suppose that the molecule has energy levels that can be described like those by a  particle on a ring. The bigger the ring the lower the energy levels become and so the lower the energy of the electronic transition. Thus incorporating a substituent that involves conjugating the electron over more atoms, increases the ring size very slightly and so  produces a longer wavelength transition.