Here, $x$ is in quadrant $II$.

i.e., $\frac{\pi}{2} < x < \pi$

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

There, $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$

are lies in first quadrant. It is given that $\tan x=-\frac{4}{3}$

$\sec ^{2} x=1+\tan ^{2} x=1+\left(\frac{-4}{3}\right)^{2}=1+\frac{16}{9}=\frac{25}{9}$

$\therefore \cos ^{2} x=\frac{9}{25}$

$\Rightarrow \cos x=\pm \frac{3}{5}$

As $x$ is in quadrant $II$, $\cos x$ is negative.

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

Now, $\cos x=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow \frac{-3}{5}=2 \cos ^{2} \frac{x}{2}-1$

$\Rightarrow 2 \cos ^{2} \frac{x}{2}=1-\frac{3}{5}$

$\Rightarrow 2 \cos ^{2} \frac{x}{2}=\frac{2}{5}$

$\Rightarrow \cos ^{2} \frac{x}{2}=\frac{1}{5}$

$ \Rightarrow \cos \frac{x}{2} = \frac{1}{{\sqrt 5 }}\quad \left[ {\because \cos \frac{x}{2}\,is\,positve} \right]$

$\therefore \cos \frac{x}{2}=\frac{\sqrt{5}}{5}$

$\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}=1$

$\Rightarrow \sin ^{2} \frac{x}{2}+\left(\frac{1}{\sqrt{5}}\right)^{2}=1$

$\Rightarrow \sin ^{2} \frac{x}{2}=1-\frac{1}{5}=\frac{4}{5}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{2}{\sqrt{5}} \quad\left[\because \sin \frac{x}{2} \text { is positive }\right]$

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

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are $\frac{2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5},$ and $2.$