It is preferred you post one question per thread.
Consider $\displaystyle -2\pi $ gives the same postition as $\displaystyle 0$ so
$\displaystyle \frac{-25\pi}{4} =\frac{(-8-8-8-1)\pi}{4}=\frac{-8\pi}{4}+\frac{-8\pi}{4}+\frac{-8\pi}{4}+\frac{-1\pi}{4}$ $\displaystyle =-2\pi-2\pi-2\pi-\frac{\pi}{4} = -\frac{\pi}{4}$
Yes, but you're still using equals signs where they shouldn't be used...
Anyway, Question 2 is correct.
In Question 3, you have $\displaystyle \sin{\theta}$ correct. For the rest you could have used the identities:
$\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1$
$\displaystyle \frac{\sin{\theta}}{\cos{\theta}} = \tan{\theta}$
$\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$
$\displaystyle \cot{\theta} = \frac{1}{\tan{\theta}}$.
If you were going to use the triangle, you have used Pythagoras' Theorem incorrectly.
It should be $\displaystyle A^2 = H^2 - O^2$. You added them.
Your Question 4 and 5 are correct.
Q. 6. (a) is correct but I would have done
$\displaystyle \csc{\frac{\pi}{3}} = \frac{1}{\sin{\frac{\pi}{3}}} = \frac{1}{\sin{60^{\circ}}}$.
(b) is correct.
Q. 7. Remember that $\displaystyle \sec^2{\frac{\pi}{6}} = \frac{1}{\cos^2{\frac{\pi}{6}}} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.
So $\displaystyle \sec^2{\frac{\pi}{3}} - 4 = \frac{4}{3} - 4 = \frac{4}{3} - \frac{12}{3} = -\frac{8}{3}$.