Step-by-step Solution:

(A) Evaluate \[ \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^{\frac{\sin x}{x - \sin x}} \] We use the standard limit result: \[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \] Also, using the first-order approximation, \[ \sin x \approx x - \frac{x^3}{6}. \] Thus, \[ x - \sin x \approx \frac{x^3}{6}. \] So, the exponent simplifies to: \[ \frac{\sin x}{x - \sin x} = \frac{x}{\frac{x^3}{6}} = \frac{6}{x^2}. \] Therefore, \[ \left( \frac{\sin x}{x} \right)^{\frac{\sin x}{x - \sin x}} = 1^{\frac{6}{x^2}} = e^0 = 1. \] (B) Evaluate \[ \lim_{x \to 0} \frac{\int_0^x t \sin^2 t \, dt}{x^2} \] Using the substitution \( f(x) = t \sin^2 t \), applying L'Hôpital's Rule, and approximating \( \sin^2 t \approx t^2 \), we get: \[ \lim_{x \to 0} \frac{\int_0^x t \sin^2 t \, dt}{x^2} = 0. \] (C) Evaluate \[ \lim_{x \to 0} (e^{2x} + x)^{\frac{1}{x}} \] Taking the natural logarithm, \[ \ln L = \frac{1}{x} \ln(e^{2x} + x). \] Approximating using \( e^{2x} \approx 1 + 2x \) and \( \ln(1+y) \approx y \), \[ \ln L = \frac{1}{x} \ln(1 + 2x + x) \approx \frac{1}{x} (2x + x) = 3. \] Thus, \[ L = e^3. \] (D) Evaluate \[ \lim_{x \to a} \frac{\log(x-a)}{\log(e^x - e^a)} \] Approximating \( e^x - e^a \approx e^a (x-a) \), \[ \log(e^x - e^a) \approx \log(e^a (x-a)) = \log e^a + \log (x-a) = a + \log (x-a). \] Thus, \[ \frac{\log(x-a)}{\log(e^x - e^a)} = \frac{\log(x-a)}{a + \log(x-a)}. \] Taking the limit as \( x \to a \), \[ \lim_{x \to a} \frac{\log(x-a)}{\log(e^x - e^a)} = e^{-1}. \] Final Matching: \[ \text{(A) } \to \text{(III) } 1, \quad \text{(B) } \to \text{(II) } 0, \quad \text{(C) } \to \text{(I) } e^3, \quad \text{(D) } \to \text{(IV) } e^{-1}. \]