Let the first term of the geometric progression be \(a\) and the common ratio be \(r\).

Given that the first term is twice the common ratio, we have:

\[a = 2r\]

The sum of an infinite geometric progression is given by the formula:

\[S_\infty = \frac{a}{1 - r}\]

Given that the sum to infinity is 8, we can write:

\[8 = \frac{2r}{1 - r}\]

Now, let's solve for \(r\):

\[8(1 - r) = 2r\]

\[8 - 8r = 2r\]

\[10r = 8\]

\[r = \frac{4}{5}\]

Substitute the value of \(r\) back into the equation \(a = 2r\):

\[a = 2\left(\frac{4}{5}\right) = \frac{8}{5}\]

The first two terms of the geometric progression are \(a\) and \(ar\):

First term: \(a = \frac{8}{5}\)

Second term: \(ar = \frac{8}{5} \times \frac{4}{5} = \frac{32}{25}\)

Now, find the sum of the first two terms:

\[S_2 = \frac{8}{5} + \frac{32}{25} = \frac{72}{25}\]