Looks easy, but it isn't!

We can rewrite the expression as:

$(\sin^{2} (5^{\circ})+\sin^{2} (85^{\circ}))+(\sin^{2} (10^{\circ})+\sin^{2} (80^{\circ}))+...+\sin^{2} (45^{\circ})+\sin^{2} (90^{\circ})$

Using $\sin(90^{\circ}-\theta)=\cos(\theta)$ :

$...=(\sin^{2} (5^{\circ})+\cos^{2} (5^{\circ}))+(\sin^{2} (10^{\circ})+\cos^{2} (10^{\circ}))+...+\frac{1}{2}+1$

Using $\sin^{2}(\theta)+\cos^{2}(\theta)=1$ :

$...=1+1+...+\frac{3}{2}$

We know there are $\frac{40-5}{5}+1=7+1=8$ pairs of sines and cosines so:

$...=8+\frac{3}{2}=\frac{19}{2}=9.5$

$\sum_{k=1}^{18}\sin^2\left(\frac{k\pi}{36}\right)=9-\frac{1}{2}\sum_{k=1}^{18}\cos\left(\frac{k\pi}{18}\right)=9-\frac{1}{2}\times(-1)=\boxed{9.5}$

We first use $\sin^2(t)=\frac{1}{2}-\frac{1}{2}\cos(2t)$ and then $\cos(t)=-\cos(\pi-t)$ .