(a) To find the maximum point \(M\), we first differentiate \(y = (x + 5) \sqrt{3 - 2x}\) using the product rule:

\(\frac{d}{dx}[(x+5)(3-2x)^{1/2}] = (x+5) \cdot \frac{1}{2}(3-2x)^{-1/2}(-2) + \sqrt{3-2x}\).

Set the derivative to zero to find critical points:

\(-(x+5)(3-2x)^{-1/2} + \sqrt{3-2x} = 0\).

Simplifying gives \((x+5) = (3-2x)\).

Solving \(x + 5 = 3 - 2x\) gives \(x = \frac{2}{3}\).

Substitute \(x = \frac{2}{3}\) back into the original equation to find \(y\):

\(y = \left( \frac{2}{3} + 5 \right) \sqrt{3 - 2 \times \frac{2}{3}} = \frac{2 \sqrt{13}}{3 \sqrt{3}}\).

Thus, the coordinates of \(M\) are \(\left( \frac{2}{3}, \frac{2 \sqrt{13}}{3 \sqrt{3}} \right)\).

(b) Using the substitution \(u = 3 - 2x\), we have \(du = -2 dx\) or \(dx = -\frac{1}{2} du\).

The limits of integration change from \(x = -5\) to \(u = 13\) and \(x = \frac{3}{2}\) to \(u = 0\).

The integral becomes:

\(\int_{13}^{0} (3-u)^{1/2} \cdot \frac{1}{2} du\).

Integrate to find the area:

\(-\frac{1}{2} \int_{13}^{0} (3-u)^{1/2} du = -\frac{1}{2} \left[ \frac{2}{3}(3-u)^{3/2} \right]_{13}^{0}\).

Evaluate the definite integral:

\(-\frac{1}{3} \left[ (3-0)^{3/2} - (3-13)^{3/2} \right] = \frac{169 \sqrt{13}}{15}\).