(a) Suppose that $f$ is analytic on the open unit disk $\{z: |z|<1 \}$ and there exists a $M$ such that $\mid f^k (0)\mid \leq k^4 M^k$ for all $k \geq 0$. Show that $f$ can be extended analytic on $\Bbb C$.
(b) Suppose that $f$ is analytic on the open unit disk $\{z: |z|<1 \}$ and there exists a $M$ such that $M>1$ and $\mid f(1/k)\mid \leq M^{-k}$ for all $k \geq 1$. Show that $f \equiv 0$.
Progress: (a) The form looks like Cauchy estimates but I have no idea which theorem in analytic continuation to use here. Generally, what kind of methods should we use to show some function can be extended analytically? (b) I'm thinking to use maximum principle. I know when $k$ goes to infinity, $f$ should go to zero.
Further progress: Since I'm not able to comment, I'll just edit here. For (a), I see your point. So we can show for each $z$, the series is always absolutely convergent and hence convergent, right? For (b), why can you write $f$ to be a multiple of such $g$?
For (a), look at the Maclaurin series for $f$: $$ \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}\,z^k. $$ Use the given estimate on $|f^{(k)}(0)|$ together with (for example) the root test to show that the series converges for all $z$. This gives you an analytic continuation of $f$ to all of $\mathbb{C}$.
(b) If $f$ is not identically zero, we can write $f(z) = z^n g(z)$ for some postive integer $n$ and an analytic function $g$ with $g(0) \neq 0$. We get $$ |f(1/k)| = \left| \frac{1}{k^n} g(1/k) \right| \le M^{-k} $$ so $$ |g(1/k)| \le \frac{k^n}{M^{k}} $$ but this tends to $0$ as $k \to \infty$ (regardless of the value of $n$). Hence $g(0) = 0$, which is a contradiction.