1. f(x)=x5+ax-1, a>0, has AT LEAST one root, since it is continuous and f(0)=-1<0 and f(1)=a. By the intermediate value theorem, f(x) must assume every value between -1 and a>0, including 0, so it has a root between 0 and 1.
f(x) has AT MOST one root, since it is differentiable and f'(x)=5x4+a, which is always positive, meaning f(x) is strictly monotonic (increasing). Monotonic functions have at most one root.
Therefore, f(x) has EXACTLY one root.
2. (a) f(x)= sin2x + cos x, 0<x<pi
f'(x) = 2sin x cos x - sin x = sin x (2cos x -1)
Set the derivative equal to zero to find the critical points:
sin x (2cos x -1) =0
sin x≠0 on 0<x<pi, which leaves us with
2cos x -1 =0, cos x = 1/2, x= pi/3
The only critical point in the interval 0<x<pi is at x = pi/3
To determine its nature look at the 2nd derivative:
f''(x) = 2cos² x -2sin² x - cos x
f''(pi/3) = -2sin²(pi/3) <0
By the 2nd derivative test, there is a maximum at x=pi/3.
Its value is f(pi/3) = sin²(pi/3) = 3/4.
I'll leave the rest of your problems to others. :)