 Solution

Yes. An explicit holomorphic surjection is the rational functionIts denominator never vanishes inside the unit disc. To prove surjectivity, take any $w\in \mathbb{C}$ and choose a square root $r$ of $w$ with $\mathrm{Re}r\ge 0$; one of the two opposite roots always has this property. Define $z=r/(1+r)$. Then $1+r\ne 0$ andso $|z|<1$. Also $z/(1-z)=r$, and consequently $F(z)=r^2=w$.

This is a holomorphic surjection from the unit disc onto the complex plane. It does not contradict part (i): it is not injective. For example, $F(i/(1+i))=F(-i/(1-i))=-1$, with two distinct arguments of modulus $1/\sqrt{2}$.