(a)
For 0 \le x < \dfrac \pi 2, 0 < \cos x \le 1, so \sec x \ge 1. So f(x) = \sec x + 2 \ge 3. So the range of f is [3, \infty).
(b)
We have:
x = \sec (f^{-1}(x)) + 2
So:
x - 2 = \sec(f^{-1}(x))
giving:
\displaystyle \frac 1 {x - 2} = \cos(f^{-1}(x))
Since 0 \le f^{-1}(x) < \dfrac \pi 2 (the domain of f), we have:
\displaystyle \arccos \left(\frac 1 {x - 2}\right) = f^{-1}(x)
The domain of f^{-1} is the range of f, so x \ge 3, or [3, \infty).
[such a function exists because f is injective]