Given that SinA = 3/5 for 0° < A < 90° and SinB = sqrt(3)/3 for 180° < B < 270°, find the exact value of tan(A - B).

Given $\sin A = \frac{3}{5}$ for $0^\circ < A < 90^\circ$ (Quadrant I). In Quadrant I, $\cos A$ is positive.
Using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:

$$\cos^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$ $$\cos A = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

Now, calculate $\tan A$:

$$\tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4}$$

Given $\sin B = \frac{\sqrt{3}}{3}$ for $180^\circ < B < 270^\circ$.
Contradiction: In the third quadrant ($180^\circ < B < 270^\circ$), the sine function is negative. However, the given value $\sin B = \frac{\sqrt{3}}{3}$ is positive. This means there is no angle $B$ that satisfies both conditions simultaneously.
Assumption: To proceed with the problem, we will assume there is a typo in the question and that $\sin B$ was intended to be negative, i.e., $\sin B = -\frac{\sqrt{3}}{3}$, which is consistent with $B$ being in the third quadrant.
With $\sin B = -\frac{\sqrt{3}}{3}$ and $B$ in Quadrant III, $\cos B$ is negative.
Using the Pythagorean identity $\sin^2 B + \cos^2 B = 1$:

$$\cos^2 B = 1 - \left(-\frac{\sqrt{3}}{3}\right)^2 = 1 - \frac{3}{9} = 1 - \frac{1}{3} = \frac{2}{3}$$ $$\cos B = -\sqrt{\frac{2}{3}} = -\frac{\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{6}}{3}$$

Now, calculate $\tan B$:

$$\tan B = \frac{\sin B}{\cos B} = \frac{-\sqrt{3}/3}{-\sqrt{6}/3} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

The formula for $\tan(A - B)$ is:

$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

Substitute the values $\tan A = \frac{3}{4}$ and $\tan B = \frac{\sqrt{2}}{2}$:

$$\tan(A - B) = \frac{\frac{3}{4} - \frac{\sqrt{2}}{2}}{1 + \left(\frac{3}{4}\right)\left(\frac{\sqrt{2}}{2}\right)}$$ $$\tan(A - B) = \frac{\frac{3}{4} - \frac{2\sqrt{2}}{4}}{1 + \frac{3\sqrt{2}}{8}}$$ $$\tan(A - B) = \frac{\frac{3 - 2\sqrt{2}}{4}}{\frac{8 + 3\sqrt{2}}{8}}$$ $$\tan(A - B) = \frac{3 - 2\sqrt{2}}{4} \times \frac{8}{8 + 3\sqrt{2}}$$ $$\tan(A - B) = \frac{2(3 - 2\sqrt{2})}{8 + 3\sqrt{2}} = \frac{6 - 4\sqrt{2}}{8 + 3\sqrt{2}}$$

To rationalize the denominator, multiply the numerator and denominator by the conjugate $8 - 3\sqrt{2}$:

$$\tan(A - B) = \frac{6 - 4\sqrt{2}}{8 + 3\sqrt{2}} \times \frac{8 - 3\sqrt{2}}{8 - 3\sqrt{2}}$$ $$\tan(A - B) = \frac{(6)(8) + (6)(-3\sqrt{2}) + (-4\sqrt{2})(8) + (-4\sqrt{2})(-3\sqrt{2})}{8^2 - (3\sqrt{2})^2}$$ $$\tan(A - B) = \frac{48 - 18\sqrt{2} - 32\sqrt{2} + 12(2)}{64 - 9(2)}$$ $$\tan(A - B) = \frac{48 - 50\sqrt{2} + 24}{64 - 18}$$ $$\tan(A - B) = \frac{72 - 50\sqrt{2}}{46}$$ $$\tan(A - B) = \frac{2(36 - 25\sqrt{2})}{2(23)}$$ $$\tan(A - B) = \boxed{\frac{36 - 25\sqrt{2}}{23}}$$