(i) The curve cuts the \(x\)-axis where \(y = 0\). For \(y = \frac{\ln x}{\sqrt{x}} = 0\), \(\ln x = 0\) which gives \(x = 1\). Thus, the coordinates of \(A\) are \((1, 0)\).

(ii) To find the maximum point \(M\), we differentiate \(y = \frac{\ln x}{\sqrt{x}}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{\ln x}{\sqrt{x}} \right) = \frac{\sqrt{x} \cdot \frac{1}{x} - \ln x \cdot \frac{1}{2}x^{-1/2}}{x} = \frac{1 - \frac{1}{2} \ln x}{x^{3/2}}\).

Setting the derivative to zero gives:

\(1 - \frac{1}{2} \ln x = 0\)

\(\ln x = 2\)

\(x = e^2\).

(iii) To find the area under the curve from \(x = 1\) to \(x = 4\), we use integration by parts:

Let \(u = \ln x\) and \(dv = \frac{1}{\sqrt{x}} \, dx\), then \(du = \frac{1}{x} \, dx\) and \(v = 2\sqrt{x}\).

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x - \int 2\sqrt{x} \cdot \frac{1}{x} \, dx\)

\(= 2\sqrt{x} \ln x - 2 \int x^{-1/2} \, dx\)

\(= 2\sqrt{x} \ln x - 4\sqrt{x}\).

Evaluating from \(x = 1\) to \(x = 4\):

\(\left[ 2\sqrt{x} \ln x - 4\sqrt{x} \right]_1^4 = (4 \ln 4 - 8) - (0 - 4)\)

\(= 4 \ln 4 - 4\)

\(= 8 \ln 2 - 4\).