Why does d/l-2,3-dibromobutane on reaction with NaI/acetone give cis-2-butene?
The process of elimination of two bromides which you are thinking is not the correct way how the reaction occurs in this case. First there will be a nucleophilic substitution (preferably SN2) by $\ce{I-}$ on the two carbon atoms containing bromine. As iodide is a better nucleophile than bromide, this substitution is majorly driven forward, and after the formation of vicinal di-iodide, the compound now undergoes bond rotations if required and make the two iodine atoms orient themselves anti-periplanar for a E2 elimination. 

So, essentially not only elimination, but substitution also occurs prior to elimination, and thus meso-2,3-dibromobutane never actually gets converted into (d/l)-2,3-dibromobutane during the reaction. Also, going by Sawhorse Projection formula is particularly useful in this case rather than Fisher projection. The former gives you a better visualisation of the syn or anti-periplanarity. 

Here is the mechanism of the reaction with the meso-compound as starting material: 



Similarly, you can draw the mechanism with the optically active 2,3-dibromobutane as your initial compound:



Now you can easily see how the corresponding alkenes are formed depending on the substrates.