(a) To solve \(\int_1^a \frac{\ln x}{\sqrt{x}} \, dx = 6\), we start by integrating:

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x - 4\sqrt{x} + C\).

Substitute the limits and equate to 6:

\(\left[ 2\sqrt{a} \ln a - 4\sqrt{a} \right] - \left[ 2\cdot1\ln 1 - 4\cdot1 \right] = 6\).

Simplify to get:

\(2\sqrt{a} \ln a - 4\sqrt{a} + 4 = 6\).

Rearrange to find \(a = \exp \left( \frac{1}{\sqrt{a}} + 2 \right)\).

(b) Calculate \(a\) for \(a = 9\) and \(a = 11\):

For \(a = 9\), \(9 < 10.31\).

For \(a = 11\), \(11 > 9.99\).

Thus, \(a\) lies between 9 and 11.

(c) Use the iterative process \(a_{n+1} = \exp \left( \frac{1}{\sqrt{a_n}} + 2 \right)\):

Start with an initial guess, say \(a_1 = 10\).

Calculate successive iterations:

\(a_2 = \exp \left( \frac{1}{\sqrt{10}} + 2 \right) \approx 10.1374\)

\(a_3 = \exp \left( \frac{1}{\sqrt{10.1374}} + 2 \right) \approx 10.1156\)

\(a_4 = \exp \left( \frac{1}{\sqrt{10.1156}} + 2 \right) \approx 10.1190\)

Continue until convergence to 2 decimal places: \(a \approx 10.12\).