5.
if some one of a,b,c is greater then 1
it is trivial
then consider a,b,c all less then 1
since a+b+c+(1-a)+(1-b)+(1-c)=3
(3/6)^6≥a(1-b)b(1-c)c(1-a)
i.e. a(1-b)b(1-c)c(1-a)≤1/64
so at least one of a(1-b),b(1-c),c(1-a) is lower or equal then 1/4
6.
let the vertices are (0,0),(a,b),(c,d) where a,b,c,d are integral numbers
(c+di)/(a+bi)=(1+i√3)/2 or (1-i√3)/2
c=(a-b√3)/2,d=(b+a√3)/2
it is possible only at a=b=c=d=0