In each case, begin with a vector equation where one side contains a linear combination of the two vectors from the span construction that gives the column space of $A$ with unknowns for scalars, and then use Theorem SLSLC to set up a system of equations. For $\vect{c}$, the resulting system has no solution, as we would expect.
For $\vect{b}$ there is a solution, as we would expect. What is interesting is that the solution is unique. This is a consequence of the linear independence of the set of two vectors in the span construction. If we wrote $\vect{b}$ as a linear combination of all four columns of $A$, then there would be infinitely many ways to do this.