Let the first term be denoted as \(a = 216\) and the common ratio as \(r\).

The fourth term is given by \(ar^3 = 64\).

Substituting the known value of \(a\), we have:

\(216r^3 = 64\)

Solving for \(r\), we get:

\(r^3 = \frac{64}{216} = \frac{8}{27}\)

\(r = \left(\frac{8}{27}\right)^{1/3} = \frac{2}{3}\)

The sum to infinity of a geometric progression is given by:

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

Substituting the values of \(a\) and \(r\), we have:

\(S_\infty = \frac{216}{1 - \frac{2}{3}} = \frac{216}{\frac{1}{3}} = 648\)

The first term of the geometric progression is \(a = 0.5\).

The common ratio \(r\) is given by \(r = \frac{0.5^3}{0.5} = 0.5^2 = 0.25\).

The formula for the sum to infinity of a geometric progression is \(S_\infty = \frac{a}{1 - r}\).

Substitute the values of \(a\) and \(r\):

\(S_\infty = \frac{0.5}{1 - 0.25} = \frac{0.5}{0.75} = \frac{2}{3}\).

Thus, the sum to infinity is \(\frac{2}{3}\) (or 0.667).

Given the first three terms of a geometric progression are 144, x, and 64.

Let the common ratio be r. Then:

\(x = 144r\)

\(64 = xr = 144r^2\)

Solving for r:

\(r^2 = \frac{64}{144} = \frac{4}{9}\)

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

Substitute r back to find x:

\(x = 144 \times \frac{2}{3} = 96\)

For the sum to infinity:

The formula for the sum to infinity of a geometric progression is:

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

where a is the first term and r is the common ratio.

Substitute the values:

\(S_\infty = \frac{144}{1 - \frac{2}{3}} = \frac{144}{\frac{1}{3}} = 432\)

Given that the grant increases by 5% each year, this forms a geometric progression (GP) with the first term, \(a = 5000\), and common ratio, \(r = 1.05\).

(i) To find the grant given in 2011, we calculate the 11th term of the GP:

\(a r^{n-1} = 5000 \times 1.05^{10}\)

\(= 5000 \times 1.62889 \approx 8144\)

(ii) To find the total amount given from 2001 to 2011, we use the sum formula for a GP:

\(S_n = a \frac{r^n - 1}{r - 1}\)

\(S_{11} = 5000 \frac{1.05^{11} - 1}{1.05 - 1}\)

\(= 5000 \frac{1.71034 - 1}{0.05}\)

\(= 5000 \times 14.2068 \approx 71034\)

The first term of the geometric progression is 81. The common ratio \(r\) can be calculated as \(r = \frac{54}{81} = \frac{2}{3}\).

The formula for the sum of the first \(n\) terms of a geometric progression is:

\(S_n = a \frac{1-r^n}{1-r}\)

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Substituting the given values:

\(S_{10} = 81 \frac{1-(\frac{2}{3})^{10}}{1-\frac{2}{3}}\)

\(S_{10} = 81 \frac{1-(\frac{2}{3})^{10}}{\frac{1}{3}}\)

\(S_{10} = 81 \times 3 \times (1-(\frac{2}{3})^{10})\)

\(S_{10} = 243 \times (1-(\frac{2}{3})^{10})\)

Calculating \((\frac{2}{3})^{10} \approx 0.01734\),

\(S_{10} = 243 \times (1-0.01734)\)

\(S_{10} = 243 \times 0.98266\)

\(S_{10} \approx 239\)