We can hydrogenate unsaturated compounds and measure the heat given off, this is called the heat of hydrogenation. As the following figure illustrates, if we hydrogenate the various butene isomers they all wind up producing the same molecule, butane.
By comparing their heats of hydrogenation we can see which butene was lowest in energy to begin with, that is, which butene is the most stable. The heats of hydrogenation reported in the figure tell us that trans-2-butene is more stable than 1-butene by about 2.3 kcal/mol. This extra stability can be attributed to the additional hyperconjugative resonance structures involving hydrogen present in trans-2-butene (6) compared to 1-butene (2).
Next let’s consider 1,3-butadiene. Here when we hydrogenate it, butane is again produced and we might estimate that it will have a heat of hydrogenation equal to twice that reported for 1-butene (1-butene is the best model because its double bond has the same substitution pattern as found in 1,3-butadiene) or 60.6 kcal/mol of heat given off. When we actually run the experiment we find that only 57.1 kcal/mol of heat is given off, 3.5 kcal/mol less than we estimated. In other words 1,3-butadiene is about 3.5 kcal/mol more stable than isolated double bonds and this is due to the resonance interaction between the two double bonds in 1,3-butadiene.
The resonance stabilization in 1,3-butadiene (3.5 kcal/mol) is larger than the hyperconjugative stabilization in trans-2-butene (2.3 kcal/mol).
Your question actually asked about 2-methyl-1,3-butadiene. In this case we have the resonance stabilization found in 1,3-butadiene plus some hyperconjugative stabilization from the additional methyl group. It will be even more stable than 1,3-butadiene. Further, 2-methyl-1,3-butadiene is more stable than any butene, so answer #1 is correct. Resonance stabilization is larger than hyperconjugative stabilization in this series of compounds.