How can I find the order of the pole z=π2 for f(z)=1(2log⁡(z))(1−sin⁡(z))? I know the...

First, note that ${\displaystyle 2\log \left(z\right)}$ does not vanish at ${\displaystyle z=\frac{\pi }{2}}$, so we can ignore it; it does not contribute to the pole. So consider simply ${\displaystyle \frac{1}{1-\sin \left(z\right)}}$. We want to write things in terms of ${\displaystyle w=z-\frac{\pi }{2}}$,
so let ${\displaystyle \sin \left(z\right)=\sin (w+\frac{\pi }{2})=\cos \left(w\right)}$, where at the end I have used a simple trig formula. Then we can expand ${\displaystyle \cos \left(w\right)=1-\frac{w^2}{2}+\frac{w^4}{24}-\cdots }$, so
${\displaystyle \frac{1}{1-\cos \left(w\right)}=\frac{1}{\frac{w^2}{2}-\frac{w^4}{24}+\cdots }=\frac{1}{w^2(\frac{1}{2}-\frac{w^2}{24}+\cdots )}}$
Thus, we can see that the order of the pole is 2.

you have the function ${\displaystyle h\left(z\right)=\log \left(z\right)(1-\sin z)}$
then ${\displaystyle h\left(\frac{\pi }{2}\right)=0}$,
${\displaystyle h^{\prime }\left(z\right)=\frac{1-\sin z}{z}-\log z\cos z\text{then}{h}^{\prime }\left(\frac{\pi }{2}\right)=0}$
${\displaystyle h{}^{″}\left(z\right)=\frac{-z\cos z-(1-\sin z)}{z^2}-\frac{\cos z}{z}+\log z\sin z\text{then}h{}^{″}\left(\frac{\pi }{2}\right)\ne 0\text{hence}\frac{\pi }{2}}$ is a zero of order two for h, therefore is a pole for ${\displaystyle \frac{1}{h\left(z\right)}}$ of order two.

Write $f(z)=\frac{1/\log z^2}{(1-\sin z)}=\frac{g(z)}{h(z)}$Observe that g(z) is analytic and non- zero at $z=\frac{\pi }{2}$ hence the order of zero of h(z) there determines the order of pole for rational function f(z).$h(\frac{\pi }{2})=h^{\prime }(\frac{\pi }{2})=0\text{and}{h}^{″}(\frac{\pi }{2})\ne 0$ suggests h(z) has zero of order 2 at $z=\frac{\pi }{2}$ so f(z) has pole of order 2 there.