There is a holomorphic function $f$ from the unit disc to itself s.t $f(0)=f(i/2)=0$. How can you prove the estimates $|f'(0)|\leq 1/2$ and $|f(-i/2)|\leq 2/5$?
Applying Schwarz's lemma we can get $|f(z)|\leq |z|$ but this gives an estimate too large. Cauchy's integral formula seems of no use since we don't know anything about other values of the function.
I'll give you a solution for the first one. Let $p = i/2$; this will simplify the notation. Let $m(z)$ be an automorphism of the unit disc, interchanging $0$ and $p$, i.e. $$m(z) = \frac{z-p}{1-\bar p z}.$$
Since $f$ maps the disc to itself and $f(0) = f(p) = 0$, we can write $$f(z) = m(z) g(z),$$ where $g$ (is holomorphic and) maps the disc to itself, satisfying $g(0) = 0$. Schwarz' lemma on $g$ gives us that $|g'(0)| \le 1$. Differentiating the equality above, we get $$ f'(z) = m'(z)g(z) + m(z)g'(z). $$ In particular $$ |f'(0)| = |m'(0)g(0) + m(0)g'(0)| \le |m'(0)g(0)| + |m(0)g'(0)| \le |m(0)| = \frac12. $$
Added: I might as well give the second half too. Again by Schwarz' lemma on $g$, $|g(z) | \le |z|$. So from the factorization above, $$\left| f\left(-\frac{i}2\right) \right| = \left|m\left(-\frac{i}2\right) g\left(-\frac{i}2\right) \right| \le \left| \frac{-i/2 - i/2}{1 - i^2/4} \right| \cdot \left| -\frac{i}2 \right| = \frac25$$
