All Tasks - Asymptote

Consider $f(x)= -2 \arctan\left(2x\right)$ with domain $D_f=\mathbb{R}$ and range $D'_f = \left] -\pi, \pi \right[$. The analytic expression of the inverse function of $f$, $f^{-1}(x)$, and $f^{-1}\left(-\dfrac{\pi}{3} \right) $ are, respectively,

Consider $f(x)= \dfrac{\pi}{2} -2 \arctan(1-2x)$ with domain $D_f=\mathbb{R}$ and let $f^{-1}(x)= \dfrac{1}{2} -\dfrac{1}{2} \tan\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$ be the analytic expression of the inverse function whose domain is $D_f^{-1}=\left]-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right[$. The solution of the equation $f\left( \dfrac{1-2x}{2}\right) +\cot\left(\arctan(1)\right) +f^{-1}(0) = 1$ is:

Consider $f(x)=-2 \arctan\left(2x\right)$ with domain $D_f=\mathbb{R}$. The solution of the equation $2\ f\left(\dfrac{x}{2} \right) -\dfrac{2}{\pi}\ f\left(\dfrac{1}{2} \right)=1$ is:

Consider $f(x)= \dfrac{\pi}{2} -2 \arctan(1-2x)$ with domain $D_f=\mathbb{R}$. The solution of the equation $f\left(\dfrac{1}{2}-x \right) + \arctan\left( \dfrac{\sqrt{3}}{3}\right)=\dfrac{\pi}{6}$ is:

Consider $f(x)= a -2 \arctan(1-2x), \ a \in \mathbb{R}$, with domain $D_f=\mathbb{R}$. Determine the analytic expression of the inverse function of $f$, $f^{-1}(x)$, and use that to calculate the value of $a \in \mathbb R$ knowing that $f^{-1}\left( -\dfrac{\pi}{2}\right) =0$.

The domain ($D_f$) and the range ($D'_f$) of the function $f(x)=- 2 \left|\dfrac{\pi}{2} -2 \arctan(1-2x)\right| $ are, respectively,