Given x = (1)/(2) and 180^ < x < 270^. This means x is in the third quadrant, where both x and x are negative.

Given $\tan x = \frac{1}{2}$ and $180^\circ < x < 270^\circ$. This means $x$ is in the third quadrant, where both $\sin x$ and $\cos x$ are negative.
We can form a right-angled triangle where the opposite side is 1 and the adjacent side is 2.
The hypotenuse $h = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$.
Therefore, in the third quadrant:

$$\sin x = -\frac{1}{\sqrt{5}}$$ $$\cos x = -\frac{2}{\sqrt{5}}$$

Using the double angle formula $\sin 2x = 2 \sin x \cos x$:

$$\sin 2x = 2 \left(-\frac{1}{\sqrt{5}}\right) \left(-\frac{2}{\sqrt{5}}\right) = 2 \left(\frac{2}{5}\right) = \frac{4}{5}$$

The expression to evaluate is $\frac{3 \sin 2x + \cos x}{\sin 2x - 2 \cos x}$.
Substitute the values of $\sin 2x$ and $\cos x$:

$$\frac{3 \left(\frac{4}{5}\right) + \left(-\frac{2}{\sqrt{5}}\right)}{\left(\frac{4}{5}\right) - 2 \left(-\frac{2}{\sqrt{5}}\right)} = \frac{\frac{12}{5} - \frac{2}{\sqrt{5}}}{\frac{4}{5} + \frac{4}{\sqrt{5}}}$$

To combine the terms in the numerator and denominator, we rationalize the $\sqrt{5}$ terms:

$$\frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \quad \text{and} \quad \frac{4}{\sqrt{5}} = \frac{4\sqrt{5}}{5}$$

Substitute these back:

$$\frac{\frac{12}{5} - \frac{2\sqrt{5}}{5}}{\frac{4}{5} + \frac{4\sqrt{5}}{5}} = \frac{\frac{12 - 2\sqrt{5}}{5}}{\frac{4 + 4\sqrt{5}}{5}} = \frac{12 - 2\sqrt{5}}{4 + 4\sqrt{5}}$$

Factor out common terms and then multiply the numerator and denominator by the conjugate of the denominator, which is $(1 - \sqrt{5})$:

$$\frac{2(6 - \sqrt{5})}{4(1 + \sqrt{5})} = \frac{6 - \sqrt{5}}{2(1 + \sqrt{5})}$$ $$\frac{6 - \sqrt{5}}{2(1 + \sqrt{5})} \times \frac{1 - \sqrt{5}}{1 - \sqrt{5}}$$

Numerator: $(6 - \sqrt{5})(1 - \sqrt{5}) = 6 - 6\sqrt{5} - \sqrt{5} + (\sqrt{5})^2 = 6 - 7\sqrt{5} + 5 = 11 - 7\sqrt{5}$
Denominator: $2(1 + \sqrt{5})(1 - \sqrt{5}) = 2(1^2 - (\sqrt{5})^2) = 2(1 - 5) = 2(-4) = -8$

$$\frac{11 - 7\sqrt{5}}{-8} = -\frac{11 - 7\sqrt{5}}{8} = \frac{-(11 - 7\sqrt{5})}{8} = \frac{-11 + 7\sqrt{5}}{8}$$ $$= \boxed{\frac{7\sqrt{5} - 11}{8}}$$