Solve the equation (3 k)2 - 3 k0.5 = 2 8.

We have $\log_2 8$. Since $8 = 2^3$, we can write:

$$\log_2 8 = \log_2 2^3 = 3$$

The term is $\log_3 k^{0.5}$.

$$\log_3 k^{0.5} = 0.5 \log_3 k$$

The original equation is $(\log_3 k)^2 - \log_3 k^{0.5} = \log_2 8$.
Substituting the simplified terms, we get:

$$(\log_3 k)^2 - 0.5 \log_3 k = 3$$

$$x^2 - 0.5x - 3 = 0$$

To eliminate the decimal, multiply the entire equation by 2:

$$2x^2 - x - 6 = 0$$

We can factor the quadratic equation. We look for two numbers that multiply to $2 \times -6 = -12$ and add to $-1$. These numbers are $-4$ and 3.

$$2x^2 - 4x + 3x - 6 = 0$$

Factor by grouping:

$$2x(x - 2) + 3(x - 2) = 0$$ $$(2x + 3)(x - 2) = 0$$

This gives two possible values for $x$:

$$2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$$ $$x - 2 = 0 \implies x = 2$$

Case 1: $x = 2$

$$\log_3 k = 2$$

Using the definition of logarithm ($b^y = x \iff \log_b x = y$):

$$k = 3^2$$ $$k = 9$$

Case 2: $x = -\frac{3}{2}$

$$\log_3 k = -\frac{3}{2}$$

Using the definition of logarithm:

$$k = 3^{-\frac{3}{2}}$$ $$k = \frac{1}{3^{\frac{3}{2}}}$$ $$k = \frac{1}{\sqrt{3^3}}$$ $$k = \frac{1}{\sqrt{27}}$$

To simplify, we can write $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$:

$$k = \frac{1}{3\sqrt{3}}$$

Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$:

$$k = \frac{1}{3\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3 \times 3} = \frac{\sqrt{3}}{9}$$

Both solutions for $k$ are positive, which is required for the logarithm $\log_3 k$ to be defined.

The values for $k$ are $\boxed{9, \frac{\sqrt{3}}{9}}$.