Step 1: Write the expression to evaluate.
We need $\sin^2\!\left(\frac{2\pi}{3}\right) + \cos^2\!\left(\frac{5\pi}{6}\right) - \tan^2\!\left(\frac{3\pi}{4}\right)$.
Step 2: Evaluate $\sin\left(\frac{2\pi}{3}\right)$.
$\frac{2\pi}{3} = \pi - \frac{\pi}{3}$, so $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$. Thus $\sin^2\!\left(\frac{2\pi}{3}\right) = \frac{3}{4}$.
Step 3: Evaluate $\cos\left(\frac{5\pi}{6}\right)$.
$\frac{5\pi}{6} = \pi - \frac{\pi}{6}$, so $\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}$. Thus $\cos^2\!\left(\frac{5\pi}{6}\right) = \frac{3}{4}$.
Step 4: Evaluate $\tan\left(\frac{3\pi}{4}\right)$.
$\frac{3\pi}{4} = \pi - \frac{\pi}{4}$, so $\tan\left(\frac{3\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$. Thus $\tan^2\!\left(\frac{3\pi}{4}\right) = 1$.
Step 5: Substitute and compute.
$\frac{3}{4} + \frac{3}{4} - 1 = \frac{6}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.
Step 6: Match with options.
The answer is $\frac{1}{2}$, which is option (2).
\[ \boxed{\dfrac{1}{2}} \]