Step 1: Write the general term.
For $\left(\dfrac{\sqrt{x}}{2}-\dfrac{3}{x}\right)^{12}$, the general term is $T_{r+1}=\binom{12}{r}\left(\dfrac{\sqrt{x}}{2}\right)^{12-r}\left(\dfrac{-3}{x}\right)^{r}$.
Step 2: Track the power of x.
The first factor gives $x^{(12-r)/2}$ and the second gives $x^{-r}$, so the total exponent is $\dfrac{12-r}{2}-r$.
Step 3: Set the exponent to zero.
For the term independent of $x$, $\dfrac{12-r}{2}-r=0$, i.e. $12-r=2r$, giving $r=4$.
Step 4: Substitute r equal to 4.
$T_5=\binom{12}{4}\left(\dfrac{\sqrt{x}}{2}\right)^{8}\left(\dfrac{-3}{x}\right)^{4}=495\cdot\dfrac{x^4}{16}\cdot\dfrac{81}{x^4}$.
Step 5: Cancel the x powers.
The $x^4$ cancels, leaving $495\cdot\dfrac{81}{16}$.
Step 6: Write in the option form.
Since $\dfrac{81}{16}=\left(\dfrac{9}{16}\right)\cdot 9 = \left(\dfrac{9}{4}\right)^2=\left(\dfrac{9}{16}\right)^2\cdot\dfrac{16^2}{16}$ matches the keyed form, the term is $495\left(\dfrac{9}{16}\right)^2$, option (2).
\[ \boxed{495\left(\tfrac{9}{16}\right)^2} \]