Let $a=\sin^{-1}\left(-\frac{1}{2}\right)$ and $b=\tan^{-1}\left(\frac{3}{4}\right)$. We have
$$\sin^2a+\cos^2a=1 \Rightarrow \frac{1}{4}+\cos^2a=1 \Rightarrow \cos a=\frac{\sqrt 3}{2} \Rightarrow \tan a = -\frac{\sqrt 3}{3}$$
since $\sin^{-1}x$ has $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ as image.
Also,
$$\frac{\sin b}{\cos b}=\frac{3}{4} \Rightarrow \frac{9}{16}\cos^2 b+\cos^2b=1 \Rightarrow \cos b=\frac{4}{5} \Rightarrow \sin b=\frac{3}{5}$$
since $\tan^{-1}x$ has $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ as image.
We then have
$$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a \tan b} = \frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3} \frac{3}{4}}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}=-\frac{4\sqrt3 + 9}{12-3\sqrt 3}\frac{12+3\sqrt 3}{12+3\sqrt 3}=$$
$$-\frac{108 + 48\sqrt 3 + 27\sqrt 3 +36}{144-27}=-\frac{144+75\sqrt 3}{117}=-\frac{48+25\sqrt 3}{39}$$
If you have angle $\theta$ in quadrant $1$, you can find its "corresponding" angle in quadrant $2$ by $(\pi - \theta)$, in quadrant $3$ by $(\pi+\theta)$, and in quadrant $4$ by $(2\pi-\theta)$. For example, $\frac{\pi}{4}$ corresponds to $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{8}$ in quadrants $2$, $3$, and $4$, respectively. (That's how I always think of them at least.)
Also, recall sine functions correspond to the height of the right triangle ($y$-axis), so they are positive in quadrants $1$ and $2$. Cosine functions correspond to base of the right triangle ($x$-axis), so they are positive in quadrants $2$ and $4$. (Tangent functions can be found through sine and cosine functions.)
You can use the following identities (which are derived from the aforementioned facts).
$$\sin\bigg(\frac{\pi}{2}+\theta\bigg) = \cos\theta \quad \sin\bigg(\frac{\pi}{2}-\theta\bigg) = \cos\theta$$
$$\cos\bigg(\frac{\pi}{2}+\theta\bigg) = -\sin\theta \quad \cos\bigg(\frac{\pi}{2}-\theta\bigg) = \sin\theta$$
$$\tan\bigg(\frac{\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(\pi+\theta\bigg) = -\sin\theta \quad \sin\bigg(\pi-\theta\bigg) = \sin\theta$$
$$\cos\bigg(\pi+\theta\bigg) = -\cos\theta \quad \cos\bigg(\pi-\theta\bigg) = -\cos\theta$$
$$\tan\bigg(\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(\pi-\theta\bigg) = -\tan\theta$$
$$\sin\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cos\theta \quad \sin\bigg(\frac{3\pi}{2}-\theta\bigg) = -\cos\theta$$
$$\cos\bigg(\frac{3\pi}{2}+\theta\bigg) = \sin\theta \quad \cos\bigg(\frac{3\pi}{2}-\theta\bigg) = -\sin\theta$$
$$\tan\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{3\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(2\pi+\theta\bigg) = \sin\theta \quad \sin\bigg(2\pi-\theta\bigg) = -\sin\theta$$
$$\cos\bigg(2\pi+\theta\bigg) = \cos\theta \quad \cos\bigg(2\pi-\theta\bigg) = \cos\theta$$
$$\tan\bigg(2\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(2\pi-\theta\bigg) = -\tan\theta$$
I certainly wouldn't recommend memorizing these though since knowing how the unit circle works basically means you know them already.
For example, in an equation you reach $$\cos \theta = -\frac{\sqrt{3}}{2}$$
You already know that $\cos {\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$ and you also know cosine is negative in quadrants $2$ and $3$, so all you need to do is find the corresponding angle for ${\frac{\pi}{6}}$ in those quadrants.
$$\text{Quadrant II} \implies \theta = \pi-{\frac{\pi}{6}} = \frac{5\pi}{6}$$
$$\text{Quadrant III} \implies \theta = \pi+{\frac{\pi}{6}} = \frac{7\pi}{6}$$
This might take a bit of practice, but once you get this whole "corresponding" angle concept, it all becomes simple. Perhaps you can start by trying to visualize this by solving equations with a unit circle. You'll eventually get the hang of it.
Best Answer
You can't directly substitute the $\frac{3\pi}{2}$ directly as the reasons you noted. However, we can take a limit. We know that $$\tan \left(\frac{17\pi}{12}\right)=\lim_{x\to \frac{18\pi}{12}} \frac{\tan x+\tan \left(-\frac{\pi}{12}\right)}{1-\tan x\tan\left(-\frac{\pi}{12}\right)}$$ Note that as $x\to \frac{18\pi}{12}$, we have $\tan x\to \infty$, so this is equivalent to $$\tan \left(\frac{17\pi}{12}\right)=\lim_{u\to \infty} \frac{u+\tan \left(-\frac{\pi}{12}\right)}{1-u\tan\left(-\frac{\pi}{12}\right)}$$ Using horizontal asymptote theorem (or you can conclude this with algebraic or other methods), we know that this is equivalent to $$\tan \left(\frac{17\pi}{12}\right)=\frac{1}{-\tan\left(-\frac{\pi}{12}\right)}$$ $$\tan \left(\frac{17\pi}{12}\right)=\frac{1}{\tan\left(\frac{\pi}{12}\right)}$$ However, this is also not too hard to prove geometrically (draw a unit circle), and is not much useful for calculating the value of $\tan\left(\frac{17\pi}{12}\right)$. The reason why we split $\frac{17\pi}{12}$ into $\frac{2\pi}{3}$ and $\frac{3\pi}{4}$ is because we already know the values of $\tan\left(\frac{2\pi}{3}\right)$ and $\tan\left(\frac{3\pi}{4}\right)$. On the other hand, using your method I'm assuming you don't already know the value of $\tan\left(\frac{\pi}{12}\right)$, so it's not as useful.