To simplify the notation, we use $f(x)$ instead of $f_c(x)$, $f^{(n)}(x)$ as $n$ th composition of $f$. The idea is similar to http://www.mathlinks.ro/viewtopic.php?t=256478. a)c) are trivial. b) For $c = 1$, we can easily show $A_1 = B_1$. Suppose $|z_0| > 1.7$. Obviously $p(z) = z$ will not hold. Let $p(z) = f^{(n)}(z)$. We can show by induction in $n$ that $|p(z_0)|\ge 1.7$. Contradiction! For d)e)f)g), it suffices to show the sequence $0,f(c),\cdots,f^{(n)}(c)$ has infinitely many distinct terms. Suppose it is not true, then there exists $n\ne m$ such that $f^{(n)}(c) = f^{(m)}(c)$. If we define $g_1(x) = x^2 + x,g_n(x)=g_{n-1}(x)^2+x$. It is equivalent to say $G(x) = g_n(x) - g_m(x) = 0$ has root $c$. Now f) is straightforward. e) is also trivial because $G(x)$ is a monic polynomial, then rational root has to be integer. For d), we define $z_0 = c^2 + c,z_n = f(z_{n - 1})$. We can prove by induction that $|z_n| = |z_{n - 1}^2 + c| > |z_{n - 1}|$ as follows: Suppose by induction assumption $|z_{n - 1}| > |c|$, then $|z_n|\ge |z_{n - 1}|^2 - |c| > |c|^2 - |c| > |c|$. Then $|z_n| - |z_{n - 1}|\ge |z_{n - 1}|^2 - |z_{n - 1}| - |c| > |c|^2 - 2|c| > 0$. For g), $c < - 2$ is covered by d). Now assume $c > \frac14$. Still by induction, $z_n - z_{n - 1} = z_{n - 1}^2 - z_{n - 1} + c > (z_{n - 1} - \frac12)^2 > 0$. Q.E.D.