C is a 3x5 matrix and {s,t,u,v,w} is a linearly independent set of vectors in R^5. Cu=0,Cv=0,Cw=0. Answer the following What are the possible values of the nullity of C?
I said the nullity must be at least 3, and the rank must be at most 2 because the null space has at least a dimension of 3.
What are the possible values for the rank of C?
See above.
Is Cx = b consistent for every choice of b in R^3?
I want to say no because I'm not given any vectors in C so I can't conclude whether or not it is consistent.
Suppose in addition that {Cs,Ct} is a linearly independent set of vectors. What would your answer to the first and second questions be now?
Nullity must be 5, therefore the rank must be 0?

You are on the right track, but you missed a few points.
(1) Note that the question is after all the possible values of $C$'s nullity, at least 3 is a bit inconcrete. For example the nullity cannot be 7 (as $C \in \def\M{\operatorname{Mat}}\M_{3,5}(\def\R{\mathbf R}\R)$, and hence $\def\nul{\operatorname{nullity}}0 \le \nul C \le 5$, together with your remark we have that 
$$ \nul C \in \{3,4,5\} $$
(2) Here you are correct, we must have $\def\r{\operatorname{rank}}\r C \in \{0,1,2\}$.
(3) Here you missed the track, note that we do not have that $Cs = Ct = 0$, but that $\{Cs, Ct\}$ is linear independent. Hence, the image of $C$ is at least two-dimensional (as it contains $Cs$ and $Ct$). This gives $\r C \ge 2$ and (1) and (2) imply $$ \nul C =3, \r C = 2$$
