Use the double angle formula for ^-1 x: 2 ^-1 x = ^-1 ((2x)/(1-x2)).

$2 \tan^{-1} x = \tan^{-1} \left(\frac{2x}{1-x^2}\right)$.
Let $x = \frac{1}{5}$.

The expression becomes $\tan \left(\tan^{-1} \frac{5}{12} - \frac{\pi}{4}\right)$.

$\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
Let $A = \tan^{-1} \frac{5}{12}$ and $B = \frac{\pi}{4}$.
Then $\tan A = \frac{5}{12}$ and $\tan B = \tan \frac{\pi}{4} = 1$.

The final answer is $\boxed{-\frac{7}{17}}$.

This implies $\cos \theta = \frac{\sqrt{5}}{3}$. Since $\cos^{-1}$ returns a value in $[0, \pi]$, $\theta$ is in the first quadrant, so $\frac{\theta}{2}$ is also in the first quadrant.

Since $\theta$ is in the first quadrant, $\sin \theta$ is positive.

$\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}$.

The final answer is $\boxed{\frac{3 - \sqrt{5}}{2}}$.