(a) To solve \(\int_{1}^{a} x^2 \ln x \, dx = 4\), we integrate by parts. Let \(u = \ln x\) and \(dv = x^2 \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^3}{3}\).

Using integration by parts, \(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \ln x - \frac{1}{9} x^3\).

Evaluating from 1 to \(a\), we have:

\(\left[ \frac{x^3}{3} \ln x - \frac{1}{9} x^3 \right]_1^a = \frac{a^3}{3} \ln a - \frac{1}{9} a^3 - \left( 0 - \frac{1}{9} \right) = \frac{a^3}{3} \ln a - \frac{1}{9} a^3 + \frac{1}{9}\).

Setting this equal to 4 gives:

\(\frac{a^3}{3} \ln a - \frac{1}{9} a^3 + \frac{1}{9} = 4\).

Multiply through by 9 to clear fractions:

\(3a^3 \ln a - a^3 + 1 = 36\).

Rearrange to find:

\(3a^3 \ln a - a^3 = 35\).

\(a^3 (3 \ln a - 1) = 35\).

Thus, \(a = \left( \frac{35}{3 \ln a - 1} \right)^{\frac{1}{3}}\).

(b) Calculate for \(a = 2.4\) and \(a = 2.8\):

For \(a = 2.4\), \(3 \ln 2.4 - 1 \approx 1.632\), \(\frac{35}{1.632} \approx 21.44\), \(21.44^{\frac{1}{3}} \approx 2.77\).

For \(a = 2.8\), \(3 \ln 2.8 - 1 \approx 2.036\), \(\frac{35}{2.036} \approx 17.19\), \(17.19^{\frac{1}{3}} \approx 2.58\).

Since \(2.4 < 2.77\) and \(2.58 < 2.8\), \(a\) lies between 2.4 and 2.8.

(c) Use the iterative formula \(a_{n+1} = \left( \frac{35}{3 \ln a_n - 1} \right)^{\frac{1}{3}}\).

Starting with \(a_1 = 2.6\):

\(a_2 = \left( \frac{35}{3 \ln 2.6 - 1} \right)^{\frac{1}{3}} \approx 2.635\).

\(a_3 = \left( \frac{35}{3 \ln 2.635 - 1} \right)^{\frac{1}{3}} \approx 2.645\).

\(a_4 = \left( \frac{35}{3 \ln 2.645 - 1} \right)^{\frac{1}{3}} \approx 2.640\).

\(a_5 = \left( \frac{35}{3 \ln 2.640 - 1} \right)^{\frac{1}{3}} \approx 2.640\).

Thus, \(a \approx 2.64\) to 2 decimal places.