Primary alkyl halides generally give SN2. The only exception would be something like neopentyl halides, which are very sterically hindered.

The terminal alkyne will deprotonate first to make the acetylide. After that, the acetylide will deprotonate 2-bromo-2methylpropane to give an alkene (E2).

Your textbook actually does a better job of describing reality than most textbooks do. In reality the SN2 is pretty tough for secondary alkyl halides/ tosylates. May I ask what textbook you are using?

Really, it’s a more quantitiative way of saying that hydroxide and alkoxides (pKa of conjugate acid around 14 and above) will tend to do eliminations, but thiolates (pka of conjugate acid around 10-11) will do substitutions.

One question I have is when you say, “Here’s a good rule of thumb: if the conjugate acid of the base/nucleophile is less than 12” are you referring to pKa = 12 ?

Start by drawing out the molecule. Your leaving group is S(CH3)2 and there should be a positive charge on the sulfur. You’re going to form a new pi bond by losing this leaving group and also a C-H bond on the carbon adjacent to the carbon bearing the sulfur. Try to draw out those potential products.