(i) The derivatives of the curves are \(2e^{2x-3}\) and \(\frac{2}{x}\). Setting these equal for parallel tangents gives:

\(2e^{2a-3} = \frac{2}{a}\)

\(e^{2a-3} = \frac{1}{a}\)

Taking natural logarithms on both sides:

\(2a - 3 = -\ln a\)

\(2a = 3 - \ln a\)

\(a = \frac{1}{2}(3 - \ln a)\)

(ii) Consider \(f(a) = a - \frac{1}{2}(3 - \ln a)\). Calculate \(f(1)\) and \(f(2)\):

\(f(1) = 1 - \frac{1}{2}(3 - \ln 1) = 1 - \frac{3}{2} = -0.5\)

\(f(2) = 2 - \frac{1}{2}(3 - \ln 2) = 2 - \frac{1}{2}(3 - 0.693) = 0.1535\)

Since \(f(1) < 0\) and \(f(2) > 0\), there is a root between 1 and 2.

(iii) Using the iterative formula \(a_{n+1} = \frac{1}{2}(3 - \ln a_n)\):

Start with \(a_1 = 1.5\).

\(a_2 = \frac{1}{2}(3 - \ln 1.5) = 1.3466\)

\(a_3 = \frac{1}{2}(3 - \ln 1.3466) = 1.3539\)

\(a_4 = \frac{1}{2}(3 - \ln 1.3539) = 1.3519\)

\(a_5 = \frac{1}{2}(3 - \ln 1.3519) = 1.3525\)

\(a_6 = \frac{1}{2}(3 - \ln 1.3525) = 1.3522\)

\(a_7 = \frac{1}{2}(3 - \ln 1.3522) = 1.3523\)

The value of \(a\) correct to 2 decimal places is 1.35.