The quotient, when $3 x^5-4 x^4+5 x^3-3 x^2+6 x-8$ is divided by $x^2+x-3$ is

If $\alpha_1, \alpha_2, \alpha_3, \alpha_4$ and $\alpha_5$ are the roots of $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$, then $\frac{1}{\alpha_1^2}+\frac{1}{\alpha_2^2}+\frac{1}{\alpha_3^2}+\frac{1}{\alpha_4^2}+\frac{1}{\alpha_5^2}$ is equal to

There were two women participating with some men in a chess tournament. Each participant played two games with the other. The number of games that the men played between themselves is 66 more than that of the men played with the women. Then, the total number of participants in the tournament is