Why mass spectra of 2-methylbutane do not show big peak at 71 coming from tertiary carbocation?
How do you know that losing a CH3 radical gives you a secondary carbocation?
Maybe it will perform a hydride shift and you land at the tertiary carbocation.
Always take this into account, because you can shoot yourself in the foot if you assume carbocations to be too stable. In EI, they rearrange all the time (e.g. benzylium &lt;-> tropylium).
However, seeing that the ion with $m/z~57$ subsequently looses $\mathrm{\text{:}CH_2}$, it seems reasonable to assume that the secondary carbocation is formed here.

The important point is the following: You propose that instead of loosing the methylradical, yielding a secondary carbocation (or a tertiary, considering the argument from above), there should be a loss of a hydrogen radical. 
However, the hydrogen radical is far less stable than the methyl radical, and I assume this is the reason for the regioselectivity, here.

Also keep in mind that you supply a lot of energy (70 eV). Hence, the induced fragmentation processes are highly endothermic, and should be under kinetic control.

If you really want to find out what is going on, you should probably have a look at McLafferty's/Turecek's classic textbook "Interpretation of Mass Spectra".