For part a)
 \\ \text{Where } R = \sqrt{(1)^2 +(-1)^2} = \sqrt{2} \\ \text{and } \alpha = tan^{-1} \frac{-1}{1} = tan^{-1}-1 = -\frac{\pi}{4} \\ \therefore sinx - cosx \Rightarrow \sqrt{2}sin(x + \frac{\pi}{4}))
For part b) it is just a normal trig graph, but I did it for you anyway:
[Sorry, i couldn't put the units in terms of pi]
And, for part c) just draw the line y = 1, approximating the x-values where the two graphs intersect, and you have the answer.