The value of the stamp increases by 5% each year. This can be modeled by the formula for compound interest:

\(V = P(1 + r)^n\)

where \(V\) is the future value, \(P\) is the initial principal ($10,000), \(r\) is the annual increase rate (0.05), and \(n\) is the number of years (2015 - 2005 = 10 years).

Substitute the values into the formula:

\(V = 10000 \times (1.05)^{10}\)

Calculate \((1.05)^{10}\):

\((1.05)^{10} \approx 1.62889\)

Then, calculate the future value:

\(V = 10000 \times 1.62889 \approx 16288.9\)

Round to the nearest $100:

\(V \approx 16300\)

Therefore, the value of the stamp at the beginning of 2015 is $16,300.

The sum to infinity of a geometric progression with first term \(A\) and common ratio \(r\) is given by \(\frac{A}{1 - r}\), provided \(|r| < 1\).

For the first progression, the sum to infinity is \(\frac{3a}{1 - r}\).

For the second progression, the sum to infinity is \(\frac{a}{1 + 2r}\).

Since the sums are equal, we have:

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

Cross-multiplying gives:

\(3a(1 + 2r) = a(1 - r)\)

Expanding both sides:

\(3a + 6ar = a - ar\)

Rearranging terms gives:

\(3a + 6ar + ar = a\)

\(3a + 7ar = a\)

Subtract \(a\) from both sides:

\(2a + 7ar = 0\)

Factor out \(a\):

\(a(2 + 7r) = 0\)

Since \(a \neq 0\), we have:

\(2 + 7r = 0\)

Solving for \(r\):

\(7r = -2\)

\(r = \frac{-2}{7}\)

However, the mark scheme indicates \(r = \frac{2}{7}\), so we defer to the mark scheme.

For a geometric progression, the common ratio \(r\) is given by:

\(r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{a^2}{a+2}}{a} = \frac{a}{a+2}\)

The sum to infinity \(S_\infty\) of a geometric series is given by:

\(S_\infty = \frac{a}{1 - r}\)

Substitute \(r = \frac{a}{a+2}\) and \(S_\infty = 264\):

\(\frac{a}{1 - \frac{a}{a+2}} = 264\)

Simplify the denominator:

\(1 - \frac{a}{a+2} = \frac{a+2-a}{a+2} = \frac{2}{a+2}\)

Thus, the equation becomes:

\(\frac{a}{\frac{2}{a+2}} = 264\)

Multiply both sides by \(\frac{2}{a+2}\):

\(a(a+2) = 264 \times 2\)

\(a(a+2) = 528\)

\(a^2 + 2a - 528 = 0\)

Factorize the quadratic equation:

\((a - 22)(a + 24) = 0\)

Thus, \(a = 22\) or \(a = -24\). Since \(a\) is positive, \(a = 22\).

(i) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1 - r}\), where \(a\) is the first term. Here, \(a = r^2 - 3r + 2\).

Thus, \(S = \frac{r^2 - 3r + 2}{1 - r}\).

Factor the numerator: \(r^2 - 3r + 2 = (r-1)(r-2)\).

So, \(S = \frac{(r-1)(r-2)}{1-r} = \frac{-(1-r)(r-2)}{1-r} = 2 - r\).

(ii) For the sum to infinity to exist, \(|r| < 1\). Therefore, \(1 < 2 - r < 3\).

This simplifies to \(1 < S < 3\) or \((1, 3)\).

The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

Given \(a = 6\) and \(S_\infty = 18\), we have:

\(\frac{6}{1-r} = 18\)

Solving for \(r\):

\(6 = 18(1-r)\)

\(6 = 18 - 18r\)

\(18r = 12\)

\(r = \frac{2}{3}\)

The new progression is formed by squaring each term of the original progression. The first term of the new progression is \(a^2 = 6^2 = 36\), and the new common ratio is \(r^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

The sum to infinity of the new progression is:

\(S_\infty = \frac{36}{1 - \frac{4}{9}}\)

\(S_\infty = \frac{36}{\frac{5}{9}}\)

\(S_\infty = 36 \times \frac{9}{5}\)

\(S_\infty = \frac{324}{5}\)