Ans Q2. The answer to problem 2 will be that all the coefficients are 0 because let the polynomial be

a0+a1x+a2x^2……….a672x^672 . If we plug 1

a0+a1+a2…….+a672=0

and( a0+a1+……..+a672)^2=a0^2+a1^2……..+a672^2+2(a0a1+a1a2+a2a3……….a671a672)

0=a0^2+a1^2…….a672^2+2(a0a1………….+a671a672)

a0^2+a1^2………+a672^2=-2(a0a1+………a671a672) (1)

aoa1……..+a671a672<0

only case possible is a0=-a1, a2=-a3………..a670=-a671 wher a1,a3…..a671 are ve integers. so we get a672 as a remainder hence the sum is not equal to 0.

SO the only case possible is that all the coefficients are 0.

If a0=-(a1+a2) for the least and a3=-(a4+a5) and so on then in (1) RHS will be negative and LHS positive which is not possible.

Ans Q1. yes

First join the edges of octagon to form a rectangle. Then make a join the four vertices to opp sides of the rectangle to form a 3d pipe like figure. Then extend the edges of the pipe to provide a space for inserting the pipe like structure formed. So when we extend the pipe like structure and insert it in the space a two holed torus will be formed.