If $\cos \theta = \dfrac{1}{2} = \dfrac{\text{adj}}{\text{hyp}}$, then $\sin \theta = \pm \dfrac{\sqrt{2^2-1^2}}{2} = \pm \dfrac{\sqrt{3}}{2}$ (the plus or minus is due to the fact that cosine is positive in quadrants 1 and 4, while sine is positive in quadrants 1 and 2). Next, use the sum of angles formula for tangent:

$\tan\left(\dfrac{\theta}{2} + 2\alpha + 8\right) = \dfrac{\tan\left(\dfrac{\theta}{2}\right) + \tan(2\alpha+8)}{1-\tan\left(\dfrac{\theta}{2}\right) \tan(2\alpha+8)}$

Next, substitute in $\tan\left(\dfrac{\theta}{2}\right) = \pm\dfrac{\sin \theta}{1+\cos \theta} = \pm \dfrac{\dfrac{\sqrt{3}}{2}}{1+\dfrac{1}{2}} = \pm \dfrac{\sqrt{3}}{3}$

Then, $\tan(2\alpha+8) = \dfrac{\tan(2\alpha) + \tan(8)}{1 - \tan(2\alpha)\tan(8)}$

$\tan(2\alpha) = \dfrac{2\tan \alpha}{1-\tan^2\alpha} = \dfrac{3}{4}$

Plugging in, you have:

$\tan(2\alpha+8) = \dfrac{3 + 4\tan(8)}{4 - 3\tan(8)}$

Then $\tan\left(\dfrac{\theta}{2} + 2\alpha + 8\right) = \dfrac{\pm \dfrac{\sqrt{3}}{3} + \dfrac{3+4\tan(8)}{4-3\tan(8)}}{1\mp \dfrac{\sqrt{3}}{3}\dfrac{3+4\tan(8)}{4-3\tan(8)}}$