a. We consider the following transformation
$$\left(x^2 - \frac{\sqrt{15}}{2}x + 1\right) \left(x^2 + \frac{\sqrt{15}}{2}x + 1\right) = x^4 - \frac{7}{4}x^2 + 1,$$
$$\left(x^{4} - \frac{7}{4}x^{2} + 1\right) \left(x^{4} + \frac{7}{4}x^{2} + 1\right) = x^{8} - \frac{17}{16}x^{4} + 1,$$
$$\left(x^{8} - \frac{17}{16}x^{4} + 1\right) \left(x^{8} + \frac{17}{16}x^{4} + 1\right) = x^{16} + \frac{223}{256}x^{8} + 1.$$
It follows that $f(x)$ is the quotient of $x^{16} + \frac{223}{256}x^8 + 1$ and
$$\left(x^2 + \frac{\sqrt{15}}{2}x + 1\right) \left(x^4 + \frac{7}{4}x^2 + 1\right) \left(x^8 + \frac{17}{16}x^4 + 1\right).$$

b. Suppose $\frac{P(x)}{Q(x)} = x^2 - \alpha x + 1$ where $P, Q$ are polynomials with non-negative coefficients. Substituting $x = 1$, we have
$$2 - \alpha = \frac{P(1)}{Q(1)} > 0 \text{ so } \alpha < 2.$$
We will prove that every real number $\alpha < 2$ satisfies the problem. Indeed, if $\alpha \le 0$ then the polynomial $f(x)$ itself is satisfied, so we can choose $P(x) = f(x)$, $Q(x) = 1$.
If $\alpha \in (0; 2)$, let us consider the multiplication
$$(x^2 - \alpha x + 1)(x^2 + \alpha x + 1) = x^4 + (2 - \alpha^2)x^2 + 1.$$
Continuing like that, we find that the coefficients of the first and last terms of the polynomial are always 1, and the middle coefficient is determined by the sequence $(u_n)$ as follows
$$\begin{cases} u_0 = \alpha, \\ u_{n+1} = 2 - u_n^2, \quad n \ge 0. \end{cases}$$
We will prove that there exists a positive term in this sequence. Suppose that for every $n \ge 1$, $u_n < 0$. Then, since $\alpha \in (0; 2)$ so by induction, we can show that $-2 < u_n < 0, \forall n \ge 1$. Note that
$$u_{n+1} - u_n = 2 - u_n - u_n^2 = (2 + u_n)(1 - u_n) > 0,$$
so $u_{n+1} - u_n > 0, \forall n \ge 1$; shows that this sequence increases. Since the sequence is bounded by 0 so it has a limit $L \in (-2; 0]$. By letting $n$ tend to infinity, we have
$$L = 2 - L^2 \text{ so } L \in \{1; -2\}.$$
This contradiction shows that there exists $n = N$ so that $u_N \ge 0$. Consider the polynomials sequence
$$f_n(x) = x^{2n+1} + u_n x^{2n} + 1$$
with $n = 1, 2, 3, \dots, N$ it is easy to see that $f(x)$ is the quotient of two polynomials $f_N(x)$ and $f_1(x)f_2(x) \dots f_{N-1}(x)$. Clearly, these polynomials have non-negative coefficients. $\square$