Step-by-step Solution:

To solve the matching problem, let's evaluate each integral in List - I and match it with the corresponding value in List - II.
(A) \( \int_{0}^{\frac{\pi}{2}} \frac{\sin^{4}x}{\sin^{4}x + \cos^{4}x} \, dx \)
We use the property of definite integrals: \[ \int_{0}^{\frac{\pi}{2}} \frac{\sin^{4}x}{\sin^{4}x + \cos^{4}x} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{4}x}{\sin^{4}x + \cos^{4}x} \, dx \] Adding these two integrals: \[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{4}x + \cos^{4}x}{\sin^{4}x + \cos^{4}x} \, dx = \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \] Thus: \[ I = \frac{\pi}{4} \] So, (A) matches with (IV).
(B) \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} \, dx \) Let \( I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}} \, dx \). Using the substitution \( x = \frac{\pi}{2} - t \), we get: \[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot t}} \, dt \] Adding the two forms of \( I \): \[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \] Thus: \[ I = \frac{\pi}{12} \] So, (B) matches with (III).
(C) \( \int_{0}^{1} x e^{x} \, dx \) Using integration by parts: \[ \int x e^{x} \, dx = x e^{x} - \int e^{x} \, dx = x e^{x} - e^{x} + C \] Evaluating from 0 to 1: \[ \int_{0}^{1} x e^{x} \, dx = \left[ x e^{x} - e^{x} \right]_{0}^{1} = (1 \cdot e^{1} - e^{1}) - (0 \cdot e^{0} - e^{0}) = 0 - (-1) = 1 \] So, (C) matches with (II).
(D) \( \int_{-1}^{1} x^{109} \cos^{88}x \, dx \) The integrand \( x^{109} \cos^{88}x \) is an **odd function** because \( x^{109} \) is odd and \( \cos^{88}x \) is even. The integral of an odd function over symmetric limits \([-a, a]\) is zero: \[ \int_{-1}^{1} x^{109} \cos^{88}x \, dx = 0 \] So, (D) matches with (I).

Thus, the correct matching is: \[ \text{A-IV, B-III, C-II, D-I} \]