Answer: \(r=\frac{2k^2}{3}\), first term \(=1\), and \(k=\pm1\).

Let the first term be \(a\) and the common ratio be \(r\). Then

\(ar^3=\frac{8k^6}{27},\)

and

\(ar^5=\frac{32k^{10}}{243}.\)

Divide the sixth-term equation by the fourth-term equation:

\(r^2=\frac{\frac{32k^{10}}{243}}{\frac{8k^6}{27}}=\frac{4k^4}{9}.\)

Since the common ratio is positive,

\(r=\frac{2k^2}{3}.\)

Now use \(ar^3=\frac{8k^6}{27}\):

\(a\left(\frac{2k^2}{3}\right)^3=\frac{8k^6}{27}.\)

So \(a=1\).

The sum to infinity is

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

Using \(a=1\) and \(S_\infty=3\),

\(\frac{1}{1-\frac{2k^2}{3}}=3.\)

Thus

\(1=3-2k^2,\)

so \(2k^2=2\), giving

\(k=\pm1.\)