Step 1: For (A)

\( y = \dfrac{1}{2 - \sin 3x} \)
Since \( \sin 3x \in [-1,1] \), we get \( 2 - \sin 3x \in [1,3] \).
Hence \( y \in \left[\tfrac{1}{3}, 1\right] \). → Matches with (III).

Step 2: For (B)

\( y = \dfrac{x^2 + x + 2}{x^2 + x + 1} = 1 + \dfrac{1}{x^2 + x + 1} \)
Since denominator is always positive, \( y > 1 \).
Minimum denominator = \(\tfrac{3}{4}\) at \(x = -\tfrac{1}{2}\).
So maximum \( y = 1 + \tfrac{1}{3/4} = \tfrac{7}{3} \).
Thus, Range = \((1, \tfrac{7}{3}] \). → Matches with (I).

Step 3: For (C)

\( y = \sin x - \cos x = \sqrt{2}\sin\!\left(x - \tfrac{\pi}{4}\right) \)
Hence, Range = \([-\sqrt{2}, \sqrt{2}] \). → Matches with (IV).

Step 4: For (D)

\( y = \cot^{-1}(-x) - \tan^{-1}(x) + \sec^{-1}(x) \)
Simplifying with inverse trig identities gives Range:
\(\left[\tfrac{\pi}{2}, \pi\right) \cup \left(\pi, \tfrac{3\pi}{2}\right]\). → Matches with (II).