$\frac{1+7i}{(2-i{)}^2}\text{ in }(r,\theta )$ form is 

$$\begin{gathered} \frac{1+7i}{(2-i{)}^2}=\frac{1+7i}{4+i^2-4i}=\frac{1+7i}{3-4i} \\ =\frac{(1+7i)(3+4i)}{3^2-4^2{i}^2}=\frac{-25+25i}{25}=-1+i \\ \begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1+i=(\sqrt{2},\pi /4)\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1+i=(\sqrt{2},\pi -\pi /4)=(\sqrt{2},3\pi /4)\end{array} \end{gathered}$$