Why is the enthalpy of combustion higher for n-butane as compared to 2-methylpropane?
Take 4 moles of carbon and 5 moles of H2. Synthesize a mole of n-butane and measure the heat evolved (125.6 kJ). Then repeat the process, but make 2-methyl propane (you'll measure 134.2 kJ). Then, with sufficient oxygen,  prepare 4 moles of CO2 and 5 moles of H2O from each of the previous two procedures and measure the heat of combustion (more for n-butane).

The total heat evolved to form the final oxidation products from carbon and H2 will be the same, but since the paths are different, the heats evolved in the two steps of each procedure will be different. n-Butane evolves less heat upon formation, therefore can (must) evolve more heat in the next step when it is oxidized to final products.