Answer
\[
\vec{w}+\vec{v}=\left\langle\frac{-\sqrt{2}+\sqrt{3}}{2}, \frac{\sqrt{2}+1}{2}\right\rangle
\]
\[
\vec{v}-\vec{w}=\left\langle\frac{\sqrt{2}+\sqrt{3}}{2}, \frac{-\sqrt{2}+1}{2}\right\rangle
\]
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
\[
\quad \sin \phi \|\vec{v}\|=b, \cos \phi \|\vec{v}\|=a
\]
$\vec{v}$ can be written as
\[
\vec{v}=\langle a, b\rangle=\langle\cos \phi, \sin \phi\rangle\|\vec{v}\|
\]
So:
\[
\begin{array}{l}
\vec{v}=\left\langle\cos 30^{\circ}, \sin 30^{\circ}\right\rangle \|\vec{v}\|=\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle \\
\vec{w}=\left\langle\cos 135^{\circ}, \sin 135^{\circ}\right\rangle\|\vec{w}\|=\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle
\end{array}
\]
The sum and the difference between these vectors are
\[
\begin{array}{l}
\vec{w}+\vec{v}=\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle+\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle=\left\langle\frac{\sqrt{3}-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2}\right\rangle \\
\vec{v}-\vec{w}=\left\langle\frac{3}{2}, \frac{1}{2}\right\rangle-\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle=\left\langle\frac{\sqrt{2}+\sqrt{3}}{2}, \frac{-\sqrt{2}+1}{2}\right\rangle
\end{array}
\]
