1.) I would try the following:
Use a Pythagorean identity and the double-angle identity for sine:
$$1-\sin(x)=\sin^2\left(\frac{x}{2} \right)-2\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)+\cos^2\left(\frac{x}{2} \right)$$
Factor as the square of a binomial:
$$1-\sin(x)=\left(\sin\left(\frac{x}{2} \right)-\cos\left(\frac{x}{2} \right) \right)$$
Apply a linear combination identity:
$$1-\sin(x)=2\sin^2\left(\frac{x}{2}-\frac{\pi}{4} \right)$$
Now the integral is:
$$I=\frac{1}{4}\int \csc^4\left(\frac{x}{2}-\frac{\pi}{4} \right)\,dx$$
Use the substitution:
$$u=\frac{x}{2}-\frac{\pi}{4}\,\therefore\,du=\frac{1}{2}dx$$
And we have:
$$I=\frac{1}{2}\int \csc^4(u)\,du$$
Using the Pythagorean identity:
$$\csc^2(\theta)=1+\cot^2(\theta)$$ we may write:
$$I=\frac{1}{2}\int \left(1+\cot^2(u) \right)\csc^2(u)\,du$$
Now, you should see a good substitution to use to finish...(Sun)