6. Answer : b) no, it does not.
a) Let $b_0 \le b_1 \le \dots \le b_{2n}$ denote the coefficients $a_0, a_1, \dots, a_{2n}$ of $p(x)$, which are arranged in non-decreasing order. Consider the permutation $c_0, c_1, \dots, c_{2n}$ of the numbers $a_0, a_1, \dots, a_{2n}$ such that $c_{2n} = b_{2n}, c_{2n-2} = b_{2n-1}, \dots, c_0 = b_n$ and $c_{2n-1} = b_{n-1}, c_{2n-3} = b_{n-2}, \dots, c_1 = b_0$. For the set of $c_0, c_1, \dots, c_{2n}$ we have
$$c_0 \ge c_1 \le c_2 \ge c_3 \le \dots \le c_{2n-2} \ge c_{2n-1} \le c_{2n}.$$
Show that the polynomial $q(x) = c_{2n}x^{2n} + c_{2n-1}x^{2n-1} + \dots + c_1x + c_0$ has no real roots. The coefficients of $q(x)$ are positive, so
$$q(x) > 0 \quad (1)$$
for $x \ge 0$. Show that inequality (1) holds for negative $x$, too. Consider two cases: 1) $-1 \le x < 0$ and 2) $x < -1$.
In case 1) we have $|x|^{2k+1} \le x^{2k}$ for $k \in \{0, 1, \dots, n-1\}$, hence, $c_{2k+1}x^{2k+1} + c_{2k}x^{2k} \ge (c_{2k} - c_{2k+1})|x|^{2k} \ge 0$. Therefore,
$$q(x) = c_{2n}x^{2n} + (c_{2n-1}x^{2n-1} + c_{2n-2}x^{2n-2}) + \dots + (c_1x + c_0) \ge c_{2n}x^{2n} > 0.$$
In case 2) we have $|x|^{2k} > |x|^{2k-1}$, hence,
$$c_{2k}x^{2k} + c_{2k-1}x^{2k-1} > (c_{2k} - c_{2k-1})x^{2k} \ge 0$$
for $k \in \{1, 2, \dots, n\}$. Therefore,
$$q(x) = (c_{2n}x^{2n} + c_{2n-1}x^{2n-1}) + \dots + (c_2x^2 + c_1x) + c_0 > c_0 > 0.$$
Thus, $q(x) > 0$ for all real $x$, so $q(x)$ has no real roots.

b) Consider the polynomial $p(x) = x^{2n} + x^{2n-1} + \dots + x - 2n$. The sum of the coefficients of $p(x)$ is equal to 0, i.e., $p(1) = 0$. Therefore, for any permutation $c_0, c_1, \dots, c_{2n}$ of the numbers $-2n, 1, \dots, 1$ we have $q(1) = 0$, where $q(x) = c_{2n}x^{2n} + c_{2n-1}x^{2n-1} + \dots + c_1x + c_0$. It follows that 1 is a root of the polynomial $q(x)$.