Step 1: Recall the log form. The inverse hyperbolic cosecant is $\text{cosech}^{-1}x=\log\!\left(\dfrac1x+\sqrt{1+\dfrac{1}{x^2}}\right)$. We apply it twice. Step 2: Evaluate the first term. For $x=2$: $\text{cosech}^{-1}2=\log\!\left(\dfrac12+\sqrt{1+\dfrac14}\right)=\log\!\left(\dfrac{1+\sqrt5}{2}\right)$. Step 3: Evaluate the second term. For $x=-\dfrac12$: $\text{cosech}^{-1}\!\left(-\dfrac12\right)=\log\!\left(-2+\sqrt5\right)$. Step 4: Add the two logs. A sum of logs is the log of the product: $\log\!\left[\dfrac{(1+\sqrt5)(\sqrt5-2)}{2}\right]$. Step 5: Multiply the bracket. Expand $(1+\sqrt5)(\sqrt5-2)=\sqrt5-2+5-2\sqrt5=3-\sqrt5$. Step 6: Write the result. So the sum is $\log\!\left(\dfrac{3-\sqrt5}{2}\right)$. \[ \boxed{\log\left(\frac{3-\sqrt5}{2}\right)} \]