I am sorry for the late response.

so when i first differentiate

i get

which is the same as

Mr F says: You ought to make life easy for yourself and recognise that 2 sinx cosx = sin(2x).
then when i differentiate again i get

Mr F says: Which (no surprise given what I said above) is the same as 2 cos(2x).
which equals 1

and when

. how do i justify that without saying i looked at my gragh?

Mr F says: Solve 2 cos(2x) = 1 ( a simple trig equation for you I hope).
b)there is no minimum in the given interval of

because when f'(x)=0 that point is a maximum on the original graph. correct?

Mr F says: There is no maximum at the value of x such that f'(x) = 0 (that is, x = 3pi/4).
c) infection points occur when

so that is when

and

are these correct?

Mr F says: They are the correct solutions to f''(x) = 0. The fact that they do or do not give points of inflection requires justification. Note that x = 3pi/4 is a solution to f'(x) = 0. It can't be a point of inflection AND a maximum turning point (as you say in Q2) .... The resolution of this is left for you to think about for the moment.