(a) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1-r}\) and \(2S = \frac{a}{1-R}\).

Equating the sums: \(\frac{a}{1-r} = \frac{2a}{1-R}\).

Cross-multiplying gives: \(a(1-R) = 2a(1-r)\).

Expanding and simplifying: \(a - aR = 2a - 2ar\).

Rearranging gives: \(2ar - aR = a\).

Dividing by \(a\) (assuming \(a \neq 0\)): \(2r - R = 1\).

Thus, \(r = 2R - 1\).

(b) The 3rd term of the first progression is \(ar^2\) and the 2nd term of the second progression is \(aR\).

Given \(ar^2 = aR\), we have \(r^2 = R\).

Substituting \(r = 2R - 1\) into \(r^2 = R\):

\((2R - 1)^2 = R\).

Expanding gives: \(4R^2 - 4R + 1 = R\).

Rearranging: \(4R^2 - 5R + 1 = 0\).

Solving this quadratic equation using the quadratic formula:

\(R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4, b = -5, c = 1\).

\(R = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8}\).

Thus, \(R = 1\) or \(R = \frac{1}{4}\).

For \(R = \frac{1}{4}\), substituting back gives \(S = \frac{a}{1-r} = \frac{a}{1-(2 \times \frac{1}{4} - 1)} = \frac{a}{\frac{1}{2}} = 2a\).

Since \(2S = 2a\), \(S = \frac{2a}{3}\).