Answer
(a) $$\operatorname{proj}_{{v}} {u} = \frac{-8}{20}(4,2).$$
(b) $$\operatorname{proj}_{{u}} {v} = \frac{-8}{5}( -1,-2).$$
Work Step by Step
Let $u=(-1,-2)$, $v=(4,2)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have $$\langle{u}, {v}\rangle=-8, \quad \langle{u}, {u}\rangle=5, \quad \langle{v}, {v}\rangle=20.$$ (a) The orthogonal projection of $u$ onto $v$ is given by $$\operatorname{proj}_{{v}} {u} =\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{-8}{20}(4,2).$$ (b) The orthogonal projection of $v$ onto $u$ is given by $$\operatorname{proj}_{{u}} {v} =\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{-8}{5}( -1,-2).$$
(c) See graph.