Answer: (i) No, (ii) Yes.

a. (i)
Let $a$, $b$, $c$ and $d > 0$ be integers and suppose that
$$F(x) = a + bx + cx^2 + dx^3 = d(x - \alpha)(x - \beta)(x - \gamma)$$
is a good polynomial with $0 < \alpha, \beta, \gamma < 1$. Let $Q(x) = x(1-x)(2x-1)$. Then it is easy to see that $|Q(x)| \le \frac{1}{6\sqrt{3}}$ for any $0 \le x \le 1$. Moreover,
$$d^3 Q(\alpha) Q(\beta) Q(\gamma) = 8F(0)F(1)F(1/2) \neq 0$$
is an integer. It follows that $d \ge 6\sqrt{3} > 10$, thus there is no good polynomial with leading coefficient $10$.

b. (ii)
Let $a$, $b$ and $c$ be integers and let $F(x) = 13x^3 - ax^2 + bx - c$. First, suppose that $F(0) = -1$ and $F(1) = 1$, then we have $c = 1$ and $a = 11 + b$. Now suppose that $F(\frac{1}{3}) > 0$ and $F(\frac{2}{3}) < 0$, then we have $\frac{47}{6} < b < \frac{55}{6}$ and it is clear that $F(x)$ is good.
Hence $F(x) = 13x^3 - 19x^2 + 8x - 1$ for $b = 8$ and $F(x) = 13x^3 - 20x^2 + 9x - 1$ for $b = 9$ are good polynomials.