Hint for (i). Write $h(z) = \frac{1}{2}(f(z) + f(-z))$ and show that $h(z)$ is of the form $k(z^2)$.
EDIT: For (ii), here is the intuition I have, and then I'll write it more nicely.
You can write $f(z) = e^{i\theta} z^2 + m(z)$, where $m(z)$ is odd. If the function $m(z)$ is not identically zero, there is a sequence $z_n$ with $|z_n| \to 1$ and $|m(z_n)| \geq \alpha$, say. By considering both $f(-z_n)$ and $f(z_n)$, you can obtain a contradiction with the bound $|f(z)| < 1$. (Once you get close enough to the boundary of the circle, you can no longer add two vectors of length $\alpha$ in opposite directions and stay within the disk.)
All right, now that you've read that, here's a version of the same thing that's a bit prettier. We have
$$\left|\frac{f(z) - f(-z)}{2}\right|^2 = \frac{1}{2}|f(z)|^2 + \frac{1}{2}|f(-z)|^2 - |h(z)|^2 \leq 1 - |h(z)|^2 = 1 - |z|^4 \xrightarrow[|z| \to 1]{} 0,$$
which shows that $f(z) - f(-z) = 0$ by the maximum modulus principle. (The first equality is the parallelogram law.)
We want to determine whether for given $a,b,c,d$, there exists a holomorphic $f\colon D \to D$ with
- $f(a) = b$, and
- $f'(c) = d$.
A typical way to attack such a problem is the Schwarz-Pick lemma, resp. its differential version
$$\frac{\lvert f'(z)\rvert}{1 - \lvert f(z)\rvert^2} \leqslant \frac{1}{1-\lvert z\rvert^2}\tag{1}$$
for $z\in D$ when $f\colon D\to D$ is holomorphic, and if we have equality at one point, then $f$ is an automorphism of $D$.
In our case, we must check whether
$$\lvert d\rvert \leqslant \frac{1 - \lvert f(c)\rvert^2}{1-\lvert c\rvert^2}\tag{2}$$
for some holomorphic $f\colon D\to D$ with $f(a) = b$. If $(2)$ doesn't hold for any such $f$, then $(1)$ tells us that no $f$ with the prescribed properties exists, and if there is such an $f$ that $(2)$ holds, we often get enough restrictions from $(2)$ that constructing a function with the desired properties or a proof that no such function exists are easier.
In case 4., $f\colon D\to D$ with $f\left(\frac{1}{2}\right) = - \frac{1}{2}$ and $f'\left(\frac{1}{4}\right) = 1$, the Schwarz-Pick lemma tells us that we must have $\left\lvert f\left(\frac{1}{4}\right)\right\rvert \geqslant \frac{1}{4}$ since the hyperbolic distance between $f\left(\frac{1}{4}\right)$ and $-\frac{1}{2}$ can be at most equal to the hyperbolic distance between $\frac{1}{4}$ and $\frac{1}{2}$. On the other hand, $(2)$ tells us that we must have $\left\lvert f\left(\frac{1}{4}\right)\right\rvert \leqslant \frac{1}{4}$ in order to have the right hand side $\geqslant 1$. The only point in $D$ satisfying both requirements is $-\frac{1}{4}$, so if an $f$ with the desired properties exists, we must have $f\left(-\frac{1}{4}\right) = -\frac{1}{4}$, and since equality holds in $(1)$ then, it follows that $f(z) = -z$. But then we have $f'\left(\frac{1}{4}\right) = -1$, so there is no holomorphic $f\colon D \to D$ with $f\left(\frac{1}{2}\right) = -\frac{1}{2}$ and $f'\left(\frac{1}{4}\right) = 1$.
For case 2., $f\colon D\to D$ with $f\left(\frac{3}{4}\right) = \frac{3}{4}$ and $f'\left(\frac{2}{3}\right) = \frac{3}{4}$, the Schwarz-Pick lemma is not as effective. From it, we obtain the bounds $$\frac{2}{3} \leqslant \left\lvert f\left(\frac{2}{3}\right)\right\rvert \leqslant \sqrt{\frac{7}{12}},$$
which don't narrow down the possibilities for $f$ much. However, with so much space to play, we suspect that such an $f$ exists. To find one, we move the fixed point of $f$ to $0$ and consider $g = T_{3/4}\circ f \circ T_{-3/4}$, where
$$T_w \colon z \mapsto \frac{z-w}{1-\overline{w}\cdot z}.$$
We want $f$ to "shrink the unit disk towards $\frac{3}{4}$", so we make the ansatz $g(z) = c\cdot z$ for some $c\in (0,1)$ which we want to determine so that $f'\left(\frac{2}{3}\right) = \frac{3}{4}$. So we try
$$f(z) = T_{-3/4}\left(c\cdot T_{3/4}(z)\right).$$
We have $T_{3/4}(2/3) = -\frac{1}{6}$, hence
$$f'\left(\frac{2}{3}\right) = T_{-3/4}'\left(-\frac{c}{6}\right)\cdot c \cdot T_{3/4}'\left(\frac{2}{3}\right).$$
Since
$$T_w'(z) = \frac{(1-\overline{w}z) +\overline{w}(z-w)}{(1-\overline{w}z)^2} = \frac{1-\lvert w\rvert^2}{(1-\overline{w}z)^2},$$
we compute
$$T_{3/4}'(2/3) = \frac{7/16}{(1/2)^2} = \frac{7}{4};\qquad T_{-3/4}'(-c/6) = \frac{7/16}{(1-c/8)^2} = \frac{28}{(8-c)^2}$$
and find that $c$ should satisfy
$$\frac{49 c}{(8-c)^2} = \frac{3}{4}.$$
Solving the quadratic equation gives the solution
$$c = \frac{122 - 14\sqrt{73}}{3} \approx 0.7946491885181928.$$
Best Answer
Let $w\in D_0(1)$, then $|w|<1<|f(z)|$ for all $z\in \partial D_0(1)$. By Rouché's Theorem, $f$ and $f-w$ have the same number of zeros in $D_0(1)$. Since $f(0)=0$, there is $z_0$ such that $f(z_0)=w$.