Possible E2 products of S-2-bromo-2-methyl-1-phenylbutane
The major products will be (E)-2-methyl-1-phenyl-1-butene and (Z)-2-methyl-1-phenyl-1-butene, formed in a 1:1 ratio. There is are reasons both for why elimination occurs specifically across the first and second positions, and as for why the product is formed in a 1:1 ratio.

The hydrogen on position 1 are called benzylic hydrogen, and are much more acidic than aliphatic hydrogen due to the resonance stabilization of the benzyl anion. This will cause these hydrogen to be removed much more quickly than the others, resulting in a majority of 1,2-unsaturated products. This is of course not to say that the other products are not formed, but they are to a lesser extent.

The products are formed in a 1:1 ratio because there are two ways for the $\ce{H}$ and $\ce{Br}$ to be anti-periplanar. This is easily seen when constructing the Newman projections of the starting material:

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Upon extraction of the hydrogen, the electron pair forms a $\pi$-bond with the adjacent carbon, ejecting the $\ce{Br-}$, and locking in the stereochemistry.