Calculating the percentage of (+)-2- butanol in the sample
A different, more logical solution gives the correct answer as well.
Since the optical rotation of (+)-2-butanol is 12.5 degrees, the optical rotation of (-)-2-butanol is -12.5 degrees. If the mole fraction of the (+) isomer is $x$ and the mole fraction of the (-) isomer is $1-x$, then
$$-2.8 = 12.5x -12.5(1-x)$$
Solving this simple linear equation gives $x = 0.388$, which when converted to mole percentage, is $38.8\%$.