Let ω=e^2π i/7.

The numbers $\omega^k$ ($k=1,\dots,6$) are the roots of $P(z)=1+z+\dots+z^6$. (A) Pairing $k$ with $7-k$ gives $\frac1{1-\omega^k}+\frac1{1-\omega^{-k}}=1$, so the sum over six terms is $3$; correct. (B) $\prod(2-\omega^k)=P(2)=2^7-1=127$; correct. (C) Expanding $|1+\omega+\omega^3|^2$, the cross terms are $\omega^{\pm1},\omega^{\pm3},\omega^{\pm2}$, so it equals $3+2(\cos\tfrac{2\pi}7+\cos\tfrac{4\pi}7+\cos\tfrac{6\pi}7)=3+2(-\tfrac12)=2$, not $3$; incorrect. (D) $|1-\omega^k|^2=2-2\cos\tfrac{2\pi k}{7}$, so the sum is $12-2(-1)=14$; correct. Hence A, B and D. Each of these facts follows from the polynomial identity $\prod_{k=1}^{6}(z-\omega^k)=1+z+\dots+z^6$ and from $\sum_{k=1}^{6}\omega^k=-1$.