1. Answer: $1+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}=\frac{2648}{2145}$
Definition of the progression: $p_{n}(2)=8 a_{n}-4\left(a_{n}+1\right)-2\left(6 a_{n}-1\right)+6=-8 a_{n}+4=-4-16 n \rightarrow a_{n}=1+2 n$

Factorization of the polynomial $p_{n}(x)$:

Checking the divisors of the free term:

$p_{n}(-2)=-8 a_{n}-4\left(a_{n}+1\right)+2\left(6 a_{n}-1\right)+6=0$, i.e., $x=-2$ is a root of the polynomial for all $n$.

$p_{n}(3)=27 a_{n}-9\left(a_{n}+1\right)+3\left(1-6 a_{n}\right)+6=0$, i.e., $x=3$ is a root of the polynomial for all $n$.

Then $p_{n}(x)=\left(a_{n} x-1\right)(x+2)(x-3)$.

The polynomials $p_{5}(x), p_{6}(x)$ and $p_{7}(x), \quad$ in addition to the roots $\quad x=-2, \quad x=3, \quad$ have the roots $\quad x=\frac{1}{a_{5}}=\frac{1}{11}, x=\frac{1}{a_{6}}=\frac{1}{13}, x=\frac{1}{a_{7}}=\frac{1}{15}$ respectively. Then the desired sum of the roots is $1+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}=\frac{2648}{2145}$