By applying the cosine theorem we get
$$\cos \alpha = \frac{|CA|^2 + |AB|^2 - |BC|^2}{2|CA| \cdot |AB|} = \frac{25 + 16 - 49}{2 \cdot 5 \cdot 4} = -\frac{1}{5},$$
and therefore
$$$$
\begin{align*}
\sin^6 \frac{\alpha}{2} + \cos^6 \frac{\alpha}{2} &= \left(\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}\right) \left(\sin^4 \frac{\alpha}{2} - \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} + \cos^4 \frac{\alpha}{2}\right) \\
&= \left(\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}\right)^2 - 3 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2} \\
&= 1 - \frac{3}{4} \sin^2 \alpha \\
&= \frac{1}{4} + \frac{3}{4} \cos^2 \alpha \\
&= \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{25} = \frac{7}{25}.
\end{align*}