A regular octagon is inscribed in a circle of radius 1 units. The product of the distances from a fixed vertex to the other seven vertices is?

The answer will be 16.
Distance from the fixed vertex to its adjacent vertex will be = 1^2 + 1^2 - 2*1*1 cos45 = 2-2^(1/2) (Where 45 degree is the angle between the lines which are drawn from the centre to fixed vertex and its adjacent vertex)
Let's number the vertices clockwise starting from the fixed vertex as 1 then others will be 2,3,4,5,6,7,8.
Now distance 12 will be equal to 18 i.e. = 2-2^(1/2)
Similarly distance 13 will be equal to 17 i.e = 2^(1/2)
14 will be equal to 16 i.e. = 2+2^(1/2)
And 15 will be equal to = 2 (Diameter of the circle)
After multiplying all the distances, you will the value of product as 16

When an n-sided polygon is inscribed in a unit circle, the product of the distance of the vertices from a fixed vertex is n.
This is easier to prove using complex numbers. If one of the vertices is at 1, then the other vertices are the nth roots of 1. The product of the diagonals will be |1-w||1-w^2|...|1-w^n|.
where w = exp(2*pi*i/n)

To prove: (1-w)(1-w^2)...(1-w^{n-1}) = n

Since (z^n - 1) = (z-1)(z^n-1 + z^n-2 +.... 1)
so, (z^n-1 + z^n-2 + ... 1) = (z-w)(z-w^2)....(z-w^n-1)
Put z = 1, so (1-w)(1-w^2)...(1-w^{n-1}) = n

So, for an octagon, the value is 8