Step-by-step Solution:

First, find the points of intersection by setting the equations equal to each other:\n\[ (x-2)^2 = -4 + 6x - x^2 \]\n\[ x^2 - 4x + 4 = -x^2 + 6x - 4 \]\n\[ 2x^2 - 10x + 8 = 0 \]\n\[ x^2 - 5x + 4 = 0 \]\n\[ (x-1)(x-4) = 0 \]\nThe curves intersect at \( x = 1 \) and \( x = 4 \).\n\nNext, find the slopes (derivatives) of the curves at these points:\nFor \( y_1 = (x-2)^2 \), \( \frac{dy_1}{dx} = 2(x-2) \).\nFor \( y_2 = -4+6x-x^2 \), \( \frac{dy_2}{dx} = 6-2x \).\n\nAt \( x = 1 \):\n\( m_1 = 2(1-2) = -2 \)\n\( m_2 = 6-2(1) = 4 \)\n\nThe angle \( \theta \) between the curves is given by:\n\[ \tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]\n\[ \tan\theta = \left| \frac{-2 - 4}{1 + (-2)(4)} \right| = \left| \frac{-6}{1 - 8} \right| = \left| \frac{-6}{-7} \right| = \frac{6}{7} \]\n\nBy symmetry, calculating at \( x = 4 \) will yield the same acute angle magnitude.\nThus, the acute angle is \( \tan^{-1}\left(\frac{6}{7}\right) \).