cscA is the inverse of the sine of A.
[tex]cscA=(\sin A)^{-1}[/tex]Since we know the value for the cosine, using the identity
[tex]\sin ^2A+\cos ^2A=1[/tex]We can calculate the value for the sine.
[tex]\begin{gathered} (-\frac{8}{\sqrt[\square]{89}})^2+\sin ^2A=1 \\ \frac{64}{89}+\sin ^2A=1 \\ \sin ^2A=1-\frac{64}{89} \\ \sin A=\pm\sqrt[]{\frac{25}{89}} \\ \sin A=\pm\frac{5}{\sqrt[]{89}} \end{gathered}[/tex]Since we know that the angle A is located at the Quadrant II, we know its sine is positive, then
[tex]\sin A=\frac{5}{\sqrt[]{89}}[/tex]Now, using our first relation
[tex]\csc A=(\frac{5}{\sqrt[]{89}})^{-1}=\frac{\sqrt[]{89}}{5}[/tex]And this is our final answer.