In each part, describe in words the set of all vectors r = (x, y) that satisfy the stated condition 
I'm kinda confused what this question wants.. I can think of a vector with coordinates 0,1 that satisfy (a), but I don't know how to incorporate that with a "fixed" vector, this is pretty confusing..
 A: To see something, I propose you to use the euclidian 2-norm. Let us denote $r_0=(x_0,y_0)$.
(a) $\|r-r_0\| = 1 \iff \|r-r_0\|^2 = 1 \iff (x-x_0)^2 + (y-y_0)^2 = 1$. This is the equation of a circle of radius $1$ centered in $(x_0,y_0)$.
Now just adapt the "=".
(b) $\|r-r_0\| < 1 \iff \|r-r_0\|^2 < 1 \iff (x-x_0)^2 + (y-y_0)^2 < 1$. All the $r$ satisfying this equation form the interior (without the boundary) of the circle described in $(a)$. Adding the circle found in $(a)$ you can see that the set of $r$ such that $\|r-r_0\| \leq 1$ is the whole circle with its boundary.
(c) $\|r-r_0\| > 1 \iff \|r-r_0\|^2 > 1 \iff (x-x_0)^2 + (y-y_0)^2 > 1$. This the whole plane except the circle every $r$ such that $\|r-r_0\| > 1$ can't be in the circle or on its boundary.
Note that you can also see it like this:
(a) all the points at a distance exactly $1$ from $r_0$.
(b) all the points at a distance at most $1$ from $r_0$.
(c) all the points at a distance greater than $1$ from $r_0$.
Maybe you could try to draw the case where the norm is the $2$-euclidian one and $r_0 = (0,0)$. It is also interesting to see that the shape of circle depends on the norm. I made you a little picture with $r_0 = (0,0)$ to see the difference of the "circle" for different norms.

All these sets are of the form $\{(x,y) \in \mathbb{R}^2:\|(x,y)\|=1\}$ with


*

*Black: $\|(x,y)\| = \|(x,y)\|_\infty := \max(|x|,|y|)$

*Blue: $\|(x,y)\| = \|(x,y)\|_3 := (|x|^3+|y|^3)^{1/3}$

*Red: $\|(x,y)\| = \|(x,y)\|_2 := (|x|^2+|y|^2)^{1/2}$

*Pink: $\|(x,y)\| = \|(x,y)\|_{3/2}:= (|x|^{3/2}+|y|^{3/2})^{2/3}$

*Orange: $\|(x,y)\| = \|(x,y)\|_1:= |x|+|y|$
