Given functions: f(x) = (2)/(3)x - 4 and g(x) = sqrt(2x). Calculate f(3), f(-2), g(-5), g(sqrt(8)). Also, calculate the images of 0 and 6 by the function f.

Substitute $x=3$ into $f(x)$.

$$f(3) = \frac{2}{3}(3) - 4$$ $$f(3) = 2 - 4$$ $$f(3) = -2$$

Substitute $x=-2$ into $f(x)$.

$$f(-2) = \frac{2}{3}(-2) - 4$$ $$f(-2) = -\frac{4}{3} - 4$$ $$f(-2) = -\frac{4}{3} - \frac{12}{3}$$ $$f(-2) = -\frac{16}{3}$$

Substitute $x=-5$ into $g(x)$.

$$g(-5) = \sqrt{2(-5)}$$ $$g(-5) = \sqrt{-10}$$

The square root of a negative number is not a real number.

Substitute $x=\sqrt{8}$ into $g(x)$.

$$g(\sqrt{8}) = \sqrt{2\sqrt{8}}$$

Simplify $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$.

$$g(\sqrt{8}) = \sqrt{2(2\sqrt{2})}$$ $$g(\sqrt{8}) = \sqrt{4\sqrt{2}}$$

We can write this as $(4\sqrt{2})^{1/2} = (2^2 \cdot 2^{1/2})^{1/2} = (2^{2.5})^{1/2} = 2^{1.25}$.
Alternatively, $\sqrt{4\sqrt{2}} = \sqrt{4} \cdot \sqrt{\sqrt{2}} = 2 \cdot (2^{1/2})^{1/2} = 2 \cdot 2^{1/4}$.

$$g(\sqrt{8}) = 2\sqrt[4]{2}$$

Summary for 1):
$f(3) = \boxed{-2}$
$f(-2) = \boxed{-\frac{16}{3}}$
$g(-5)$ is not a real number.
$g(\sqrt{8}) = \boxed{2\sqrt[4]{2}}$

2) Calculate the images of 0 and 6 by the function $f$.

Substitute $x=0$ into $f(x)$.

$$f(0) = \frac{2}{3}(0) - 4$$ $$f(0) = 0 - 4$$ $$f(0) = -4$$

Substitute $x=6$ into $f(x)$.

$$f(6) = \frac{2}{3}(6) - 4$$ $$f(6) = 4 - 4$$ $$f(6) = 0$$

Summary for 2):
The image of 0 by $f$ is $\boxed{-4}$.
The image of 6 by $f$ is $\boxed{0}$.

3) Calculate the images of 0 and $-\sqrt{2}$ by the function $g$.

Substitute $x=0$ into $g(x)$.

$$g(0) = \sqrt{2(0)}$$ $$g(0) = \sqrt{0}$$ $$g(0) = 0$$

Substitute $x=-\sqrt{2}$ into $g(x)$.

$$g(-\sqrt{2}) = \sqrt{2(-\sqrt{2})}$$ $$g(-\sqrt{2}) = \sqrt{-2\sqrt{2}}$$

The square root of a negative number is not a real number.

Summary for 3):
The image of 0 by $g$ is $\boxed{0}$.
The image of $-\sqrt{2}$ by $g$ is not a real number.

4) Calculate the number which has 2 as its image by the function $f$.
This means finding $x$ such that $f(x) = 2$.

$$\frac{2}{3}x - 4 = 2$$

$$\frac{2}{3}x = 2 + 4$$ $$\frac{2}{3}x = 6$$

$$x = 6 \times \frac{3}{2}$$ $$x = 3 \times 3$$ $$x = 9$$

The number is $\boxed{9}$.

5) What is the antecedent of 5 by the function $g$?
This means finding $x$ such that $g(x) = 5$.

$$\sqrt{2x} = 5$$

$$(\sqrt{2x})^2 = 5^2$$ $$2x = 25$$

$$x = \frac{25}{2}$$

The antecedent is $\boxed{\frac{25}{2}}$.