(a) To find the \(x\)-coordinates of \(A\) and \(B\), solve the simultaneous equations:

\(2y = x + 5\)

\(y = x^2 - 4x + 7\)

Substitute \(y = \frac{x + 5}{2}\) into the quadratic equation:

\(\frac{x + 5}{2} = x^2 - 4x + 7\)

Multiply through by 2 to eliminate the fraction:

\(x + 5 = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

\(2x^2 - 9x + 9 = 0\)

Factorize or use the quadratic formula to solve:

\(x = 3 \text{ or } x = \frac{1}{2}\)

(b) To find the equation of the tangent to the curve at \(B\), first find the derivative:

\(\frac{dy}{dx} = 2x - 4\)

At \(x = 3\), the gradient \(m\) is:

\(m = 2(3) - 4 = 2\)

The equation of the tangent at \(x = 3\) is:

\(y - 4 = 2(x - 3)\)

(c) To find the acute angle between the tangent and the line \(2y = x + 5\), first find the gradients:

The gradient of the line \(2y = x + 5\) is \(\frac{1}{2}\).

The angle \(\theta_1\) with the \(x\)-axis for the tangent is:

\(\arctan(2) \approx 63.4^\circ\)

The angle \(\theta_2\) with the \(x\)-axis for the line is:

\(\arctan\left(\frac{1}{2}\right) \approx 26.6^\circ\)

The acute angle between them is:

\(63.4^\circ - 26.6^\circ = 36.8^\circ\)

Rounded to 1 decimal place, the angle is \(37^\circ\).