Critical values are points where the derivative is equal to zero or undefined, so:
1) f(t) = 2 cost + sin 2t
You messed up your derivative. The derivative on cos(t) is -sing(t). Since 2 is a constant it comes along for the ride. for sin(2t) you can use the chain rule:
f '(t) = -2 sin(t) + cos(2t) (2)
Now set this equal to zero:
0 = -2sin(t) +2cos(2t)
0 = -sin(t) + cos(2t)
Using a trig identity we have:
0 = -sin(t) + [1 - 2(sin(t))^2]
0 = 2(sin(t))^2 + sin(t) -1
0 = (2sin(t) -1)(sin(t) +1)
0 = 2sin(t) - 1 ==> sin(t) = 1/2 <==> t = pi/6
0 = sin(t)+1 ==> sin(t) = -1 which also does not happen in your interval, so we can ignore this point as well.
So now you have to evaluate 0, pi/6 and pi/2 to find the absolute max and min.
2. f(t) = x - ln x , [1/2, 2]
f '(t) = 1 - (1/x)
Setting this equal to zero:
0 = 1- (1/x)
(1/x) = 1
(1) = x
Now you evaluate 1/2, 1 and 2 to find the absolute max and minimum.