Step 1: Recall what angle between curves means.
The angle between two curves at a point where they meet is simply the angle between their tangent lines there. So we need the slopes of both curves at a common point.
Step 2: Find where the curves meet.
Set $(x-2)^2=-4+6x-x^2$. Expanding the left side, $x^2-4x+4=-4+6x-x^2$, so $2x^2-10x+8=0$, i.e. $x^2-5x+4=0$, giving $(x-1)(x-4)=0$. Thus $x=1$ or $x=4$.
Step 3: Differentiate the first curve.
For $y=(x-2)^2$ we have $\dfrac{dy}{dx}=2(x-2)$.
Step 4: Differentiate the second curve.
For $y=-4+6x-x^2$ we have $\dfrac{dy}{dx}=6-2x$.
Step 5: Evaluate the slopes at $x=1$.
The first slope is $m_1=2(1-2)=-2$ and the second is $m_2=6-2(1)=4$.
Step 6: Apply the angle formula.
\[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\left|\frac{-2-4}{1+(-2)(4)}\right|=\left|\frac{-6}{-7}\right|=\frac67. \] So the acute angle is $\tan^{-1}\frac67$ (the point $x=4$ gives the same acute angle).
\[ \boxed{\tan^{-1}\frac67} \]