1. The problem asks to evaluate the expression a) $\tan 2x$ for $\pi \leq x \leq \frac{3\pi}{2}$ given $\cos x = \frac{2}{45}$.\n\n2. Recall the double angle formula for tangent: $$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}.$$\n\n3. We need to find $\tan x$ first. Since $\cos x = \frac{2}{45}$, use the Pythagorean identity: $$\sin^2 x = 1 - \cos^2 x = 1 - \left(\frac{2}{45}\right)^2 = 1 - \frac{4}{2025} = \frac{2021}{2025}.$$\n\n4. Then, $$\sin x = \pm \sqrt{\frac{2021}{2025}} = \pm \frac{\sqrt{2021}}{45}.$$\n\n5. Since $x$ is in the interval $[\pi, \frac{3\pi}{2}]$, which is the third quadrant, both sine and cosine are negative. So, $$\sin x = -\frac{\sqrt{2021}}{45}.$$\n\n6. Calculate $\tan x$: $$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{2021}}{45}}{\frac{2}{45}} = -\frac{\sqrt{2021}}{2}.$$\n\n7. Substitute $\tan x$ into the double angle formula: $$\tan 2x = \frac{2 \left(-\frac{\sqrt{2021}}{2}\right)}{1 - \left(-\frac{\sqrt{2021}}{2}\right)^2} = \frac{-\sqrt{2021}}{1 - \frac{2021}{4}}.$$\n\n8. Simplify the denominator: $$1 - \frac{2021}{4} = \frac{4}{4} - \frac{2021}{4} = \frac{4 - 2021}{4} = \frac{-2017}{4}.$$\n\n9. So, $$\tan 2x = \frac{-\sqrt{2021}}{\frac{-2017}{4}} = -\sqrt{2021} \times \frac{4}{-2017} = \frac{4 \sqrt{2021}}{2017}.$$\n\n10. Final answer: $$\boxed{\tan 2x = \frac{4 \sqrt{2021}}{2017}}.$$