What are the common products of the reaction: 2-bromobutane and NaOH I am self-studying organic chemistry to better my understanding, I cannot formally enroll at a university due to being active duty. 

From my understanding there are 2 possible mechanisms when 2-bromobutane reacts with NaOH or any strong hydroxyl base: The E2 and SN2. The carbon attached to bromine is secondary. In the E2 reaction, the oxygen from the hydroxide attacks the "beta" hydrogen, deprotinating it and allowing the electron to travel and form a double bond. Bromide is a leaving group. We get 2-butene as well as water and the bromide ion. 

But we have two products from this reaction: trans-2-butene and cis-2-butene. 

We also have (2,S)-butanol. 

My question is which of these are more favored and why? I know that cis-2-butene would not be favored due to sterics. If possible, I would also like to know when E2 reactions are favored over SN2.
A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them.

An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution ($\mathrm{S_N1}$ and $\mathrm{S_N2}$) and elimination.

Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions. 

Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.
Similarly, a primary alkyl halide will prefer a $\mathrm{S_N2}$ reaction, a secondary halide – $\mathrm{S_N2}$ or elimination depending upon the strength of base/nucleophile and a tertiary halide – $\mathrm{S_N1}$ or elimination depending upon the stability of carbocation or the more substituted alkene.

So, I think preferably in aqueous NaOH or KOH there should be formation of alcohol but that too by $\mathrm{S_N1}$ if there's a tertiary; and, if primary halide, then it would follow $\mathrm{S_N2}$. It's bit tricky for secondary; again it depends on the reagents and solvents used.