1. Interpret the complex numbers

Given $a_k=\cos\left(\frac{k\pi}{7}\right)+i\sin\left(\frac{k\pi}{7}\right)=e^{ik\pi/7}.$

So each $a_k$ lies on the unit circle.

2. Use the distance formula on the unit circle

For $z_1=e^{i\alpha},\qquad z_2=e^{i\beta},$ we have $|z_1-z_2|=2\left|\sin\frac{\alpha-\beta}{2}\right|.$

This will be used repeatedly.

3. Evaluate the numerator

The numerator is $\sum_{k=1}^{12}|a_{k+1}-a_k|.$

Now $a_{k+1}=e^{i(k+1)\pi/7},\qquad a_k=e^{ik\pi/7}.$ Their arguments differ by $\frac{(k+1)\pi}{7}-\frac{k\pi}{7}=\frac{\pi}{7}.$

Hence each term is the same: $|a_{k+1}-a_k|=2\sin\left(\frac{1}{2}\cdot\frac{\pi}{7}\right)=2\sin\frac{\pi}{14}.$

There are $12$ such terms, so $\sum_{k=1}^{12}|a_{k+1}-a_k|=12\cdot 2\sin\frac{\pi}{14}=24\sin\frac{\pi}{14}.$

4. Evaluate the denominator

The denominator is $\sum_{k=1}^{3}|a_{4k-1}-a_{4k-2}|.$

For each $k$, $a_{4k-1}=e^{i(4k-1)\pi/7},\qquad a_{4k-2}=e^{i(4k-2)\pi/7}.$ Again, the difference in arguments is $\frac{(4k-1)\pi}{7}-\frac{(4k-2)\pi}{7}=\frac{\pi}{7}.$

So each term is again $|a_{4k-1}-a_{4k-2}|=2\sin\frac{\pi}{14}.$

There are $3$ such terms, hence $\sum_{k=1}^{3}|a_{4k-1}-a_{4k-2}|=3\cdot 2\sin\frac{\pi}{14}=6\sin\frac{\pi}{14}.$

5. Compute the ratio

Therefore,