Step-by-step Solution:

To match List - I with List - II, let's analyze each item step by step:
(A) Eccentricity of the conic \( x^{2} - 4x + 4y + 4y^{2} = 12 \)
First, rewrite the equation in standard form: \[ x^{2} - 4x + 4y^{2} + 4y = 12 \] Complete the square for both \( x \) and \( y \): \[ (x^{2} - 4x + 4) + 4(y^{2} + y + \frac{1}{4}) = 12 + 4 + 1 \] \[ (x - 2)^{2} + 4(y + \frac{1}{2})^{2} = 17 \] Divide by 17 to get the standard form of an ellipse: \[ \frac{(x - 2)^{2}}{17} + \frac{(y + \frac{1}{2})^{2}}{\frac{17}{4}} = 1 \] For an ellipse, eccentricity \( e \) is given by: \[ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \] Here, \( a^{2} = 17 \) and \( b^{2} = \frac{17}{4} \): \[ e = \sqrt{1 - \frac{\frac{17}{4}}{17}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] So, (A) matches with (IV).
(B) Latus rectum of conic \( 5x^{2} + 9y^{2} = 45 \)
Rewrite the equation in standard form: \[ \frac{x^{2}}{9} + \frac{y^{2}}{5} = 1 \] For an ellipse, the latus rectum \( L \) is given by: \[ L = \frac{2b^{2}}{a} \] Here, \( a^{2} = 9 \) and \( b^{2} = 5 \): \[ L = \frac{2 \times 5}{3} = \frac{10}{3} \] So, (B) matches with (I).
(C) The straight line \( x + y = a \) touches the curve \( y = x - x^{2} \), then the value of \( a \) is
For the line to be tangent to the curve, the system of equations should have exactly one solution. Substitute \( y = a - x \) into the curve equation: \[ a - x = x - x^{2} \] \[ x^{2} - 2x + a = 0 \] For the quadratic equation to have exactly one solution, the discriminant \( D \) must be zero: \[ D = (-2)^{2} - 4 \times 1 \times a = 4 - 4a = 0 \implies a = 1 \] So, (C) matches with (II).
(D) Eccentricity of conic \( 3x^{2} - y^{2} = 4 \)
Rewrite the equation in standard form: \[ \frac{x^{2}}{\frac{4}{3}} - \frac{y^{2}}{4} = 1 \] For a hyperbola, eccentricity \( e \) is given by: \[ e = \sqrt{1 + \frac{b^{2}}{a^{2}}} \] Here, \( a^{2} = \frac{4}{3} \) and \( b^{2} = 4 \): \[ e = \sqrt{1 + \frac{4}{\frac{4}{3}}} = \sqrt{1 + 3} = \sqrt{4} = 2 \] So, (D) matches with (III).
Final Matching:
- (A) matches with (IV)
- (B) matches with (I)
- (C) matches with (II)
- (D) matches with (III)
Thus, the correct matching is: \[ \boxed{A-IV, B-I, C-II, D-III} \]