$$r(t) = \left(\frac{t}{2}+\frac{1}{2}\sin(t),\frac{1}{2}\cos(t),2\cos\left(\frac{t}{2}\right)\right)$$
I worked out $$T' = \left(-\frac{1}{2}\sin(t),-\frac{1}{2}\cos(t),-\frac{1}{2}\cos\left(\frac{t}{2}\right)\right)$$ and $$\left|T'\right|=\sqrt{\frac{1}{4}+\frac{1}{4}\cos^2\left(\frac{t}{2}\right)}$$. But how do I find the normal vector and thus the binormal vector from this if I cant further simply the bit inside the roots, or have I made a mistake here.
If ${\bf r}(t)$ represents a point moving along a curve at varying of $t$, then the three infinitesimally near points $$ {\bf r}(t),\;{\bf r}(t + dt),\;{\bf r}\left( {\left( {t + dt} \right) + dt'} \right) $$ define the plane that "contains" the curve at time $t$.
That means that such a plane is parallel to the two vectors $$ \mathop {\bf r}\limits^ \bullet (t),\quad \mathop {\bf r}\limits^ \bullet (t + dt) = \mathop {\bf r}\limits^{ \bullet \bullet } (t)dt \propto \;\mathop {\bf r}\limits^{ \bullet \bullet } (t) $$ and thus that the normal $\bf m$ to that plane will be proportional to their cross product $$ {\bf m}(t) \propto \;\mathop {\bf r}\limits^ \bullet (t) \times \mathop {\bf r}\limits^{ \bullet \bullet } (t) $$
You can then find the normal to the curve in the plane, $\bf n$, by taking again the cross product $$ {\bf n}(t) \propto {\bf m}(t) \times \;\mathop {\bf r}\limits^ \bullet (t) $$ or otherwise splitting the components of the acceleration into the tangential (parallel to speed) and normal (centripetal/centrifugal).
In the above version I took the reference $(\mathop {\bf r}\limits^ \bullet, \; \bf n , \; \bf m)$ as right handed, which corresponds to (velocity, normal (centrip.) acceleration, angular velocity).