Step 1: Find $\cos x$.

Since $\tan x = -\frac{4}{3}$ and $x$ lies in quadrant II, we have:

$$\sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9}$$

Therefore $\sec x = -\frac{5}{3}$ (negative in quadrant II), giving:

$$\cos x = -\frac{3}{5}$$

Step 2: Determine quadrant of $x/2$.

Since $x$ lies in quadrant II, $\frac{\pi}{2} < x < \pi$, which means:

$$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$$

Thus $x/2$ lies in quadrant I where $\sin$, $\cos$, and $\tan$ are all positive.

Step 3: Find $\sin\left(\frac{x}{2}\right)$.

Using:

$$2\sin^2\left(\frac{x}{2}\right) = 1 - \cos x = 1 - \left(-\frac{3}{5}\right) = \frac{8}{5}$$

we get:

$$\sin^2\left(\frac{x}{2}\right) = \frac{4}{5}$$

Therefore:

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

(positive in quadrant I)

Step 4: Find $\cos\left(\frac{x}{2}\right)$.

Using:

$$2\cos^2\left(\frac{x}{2}\right) = 1 + \cos x = 1 + \left(-\frac{3}{5}\right) = \frac{2}{5}$$

we get:

$$\cos^2\left(\frac{x}{2}\right) = \frac{1}{5}$$

Therefore:

$$\cos\left(\frac{x}{2}\right) = \frac{1}{\sqrt{5}}$$

(positive in quadrant I)

Step 5: Find $\tan\left(\frac{x}{2}\right)$.

$$\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = \mathbf{2}$$