Answer
(a) $$\operatorname{proj}_{{v}} {u} =\frac{-5}{2}(0,-1,1).$$
(b) $$\operatorname{proj}_{{u}} {v} =\frac{-5}{14}( 1,3,-2).$$
Work Step by Step
Let $u=(1,3,-2)$, $v=(0,-1,1)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have $$\langle{u}, {v}\rangle=-5, \quad \langle{u}, {u}\rangle=14, \quad \langle{v}, {v}\rangle=2.$$
(a) The orthogonal projection of $u$ onto $v$ is given by $$\operatorname{proj}_{{v}} {u} =\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{-5}{2}(0,-1,1).$$
(b) The orthogonal projection of $v$ onto $u$ is given by $$\operatorname{proj}_{{u}} {v} =\frac{\langle{u}, {v}\rangle}{\langle{u}, {u}\rangle} {u}=\frac{-5}{14}( 1,3,-2).$$