Answer
\[
\begin{array}{l}
-\vec{w}+\vec{v}=\left\langle\frac{\sqrt{2}+\sqrt{3}}{2}, \frac{-\sqrt{2}+1}{2}\right\rangle \\
\vec{w}+\vec{v}=\left\langle\frac{-\sqrt{2}+\sqrt{3}}{2}, \frac{\sqrt{2}+1}{2}\right.\rangle
\end{array}
\]
Work Step by Step
Let $\vec{v}=\langle a, b\rangle$ be a vector in 2 -space. The norm of this vector and the angle with respect to the positive $x$ axis are
\[
\begin{aligned}
&\sqrt{b^{2}+a^{2}}=\|\vec{v}\| \\
b / a=\tan \phi &
\end{aligned}
\]
respectively. So
\[
\quad \|\vec{v}\| \sin \phi=b, \|\vec{v}\| \cos \phi=a
\]
and $\vec{v}$ can be written as
\[
\vec{v}=\langle a, b\rangle=\|\vec{v}\|\langle\cos \phi, \sin \phi\rangle
\]
So:
\[
\begin{array}{l}
\vec{v}=\|\vec{v}\|\langle\cos (\pi / \phi), \sin (\pi / 6)\rangle=\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle \\
\vec{w}=\|\vec{w}\|\left\langle\cos \left(\frac{3 \pi}{4}\right), \sin \left(\frac{3 \pi}{4}\right)\right\rangle=\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle
\end{array}
\]
The components of $-\vec{w}+\vec{v}$ and $\vec{w}+\vec{v}$ are
\[
\begin{array}{l}
\vec{w}+\vec{v}=\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle+\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle=\left\langle\frac{-\sqrt{2}+\sqrt{3}}{2}, \frac{\sqrt{2}+1}{2}\right\rangle \\
-\vec{w}+\vec{v}=-\left\langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle+\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle=\left\langle\frac{\sqrt{2}+\sqrt{3}}{2}, \frac{-\sqrt{2}+1}{2}\right\rangle
\end{array}
\]
