Answer
\[
\vec{w}+\vec{v}=\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle
\]
\[
\vec{v}-\vec{w}=\left\langle-\frac{3}{2}, \frac{\sqrt{3}}{2}\right\rangle
\]
Work Step by Step
Let $\vec{v}=\langle a, b\rangle$ be a vector in 2d-space. The base of this vector and angle for the positive $x$ axis are:
\[
\begin{aligned}
&\sqrt{b^{2}+a^{2}}=\|\vec{v}\| \\
b / a=\tan \phi &
\end{aligned}
\]
respectively. So
\[
\quad b=\|\vec{v}\| \sin \phi, a=\|\vec{v}\| \cos \phi
\]
$\vec{v}$ can be written as
\[
\vec{v}=\langle a, b\rangle=\langle\cos \phi, \sin \phi\rangle\|\vec{v}\|
\]
So:
\[
\begin{array}{l}
\vec{u}=\left\langle\cos 120^{\circ}, \sin 120^{\circ}\right\rangle\|\vec{v}\|=\left\langle-\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle \\
\vec{w}=\left\langle\cos 0^{\circ}, \sin 0^{\circ}\right\rangle\|\vec{w}\|=\langle 1,0\rangle
\end{array}
\]
The sum and the difference between these vectors are
\[
\begin{array}{l}
\vec{w}+\vec{v}=\langle 1,0\rangle+\left\langle-\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle=\left\langle\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle \\
-\vec{w}+\vec{v}=-\langle 1,0\rangle+\left\langle-\frac{1}{2}, \frac{\sqrt{3}}{2}\right\rangle=\left\langle-\frac{3}{2}, \frac{\sqrt{3}}{2}\right\rangle
\end{array}
\]
