Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact.

To obtain the estimate $|f'(z)|<1$, apply Cauchy theorem: $$|f'(z)|=\left|\frac{1}{2\pi}\int_{|\zeta-z|=1}\frac{f(\zeta) d\zeta}{(\zeta-z)^2}\right|\leq 1.$$ Equality in the right inequality can happen only if $f$ is constant, so we always have a strict inequality.

Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact.

To obtain the estimate $|f'(z)|<1$, apply Cauchy theorem: $$|f'(z)|=\left|\frac{1}{2\pi}\int_{|\zeta-z|=1}\frac{f(\zeta) d\zeta}{(\zeta-z)^2}\right|\leq 1.$$ Equality in the right inequality can happen only if $f$ is constant, so we always have a strict inequality.

So this estimate is not exact. To obtain the exact one, follow @Christian Remling's suggestion and use the Schwarz lemma. For convenience, let $g(z)=f(2z)$, then $g$ maps the unit disk into itself, so $$|g'(z)|\leq \frac{1-|g(z)|^2}{1-|z|^2}\leq\frac{1}{1-|z|^2}\leq \frac{4}{3}.$$ Thus the exact estimate us $|f'(z)|=|g'(z)|/2\leq 2/3.$ This is attained on a Mobius map.

Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact.

To obtain the estimate $|f'(z)|<1$, apply Cauchy theorem: $$|f'(z)|=\left|\frac{1}{2\pi}\int_{|\zeta-z|=1}\frac{f(\zeta) d\zeta}{(\zeta-z)^2}\right|\leq 1.$$

Yes, this is true. The simple reason is that bounded functions form a normal family. Therefore their derivatives are uniformly bounded on every compact.

To obtain the estimate $|f'(z)|<1$, apply Cauchy theorem: $$|f'(z)|=\left|\frac{1}{2\pi}\int_{|\zeta-z|=1}\frac{f(\zeta) d\zeta}{(\zeta-z)^2}\right|\leq 1.$$ Equality in the right inequality can happen only if $f$ is constant, so we always have a strict inequality.