Answer
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
= \frac{4}{10}(3,1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
= \frac{4}{8}( 2,-2).$$
Work Step by Step
Let $u=(2,-2)$, $v=(3,1)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have
$$\langle{u}, {v}\rangle=4, \quad \langle{u}, {u}\rangle=8, \quad \langle{v}, {v}\rangle=10.$$
(a) The orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{4}{10}(3,1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by
$$\operatorname{proj}_{{u}} {v}
=\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{4}{8}( 2,-2).$$
(c)