a) No relatively acidic protons so you won't get an aldol. On the face of it, it is tempting to say that the Cannizzaro is fine. However, have a look at the tetrahedral intermediate you get from addition of hydroxide. You have 3 possible leaving groups: 1. -OH; 2. H-; 3. -CCl3.
1. Is an equilibrium that gets you back to where you started, no net reaction.
2. Is hydride transfer, Cannizzaro reaction.
3. This is a relatively stable carbanion, a better leaving group than hydride, and is the final stage of the haloform reaction.
I would predict a haloform reaction.
b) There is a relatively acidic alpha-hydrogen, so you will get aldehyde-enolate equilibrium. There is a possibility for aldol reactions, but as you say there are problems caused by hinderance and possible inductive effects. Cannizzaro is also possible, and in this case is fine, since your choice of leaving group form the tetrahedral intermediate is hydroxide, hydride (Cannizarro), or Me2HC- (which is a truly awful leaving group). I think the major reaction here is Cannizzaro.
A nice question, since normally it is drummed into students that non-enolisable aldehydes go by Cannizzaro, and the enolisable do aldols. The Cannizzaro reaction is not limited to non-enolisable aldehydes though, and in this case the non-enolisable aldehyde is set up for a haloform reaction instead of a Cannizzaro. Cunning question.