The answer to Question 2 is no. The essential reason is the following: Du Bois - Reymond constructed a continuous function $f$ on the unit circle whose Fourier series, $\sum_n a_n e^{i n \theta}$, does not converge at $\theta=0$. If you look at the construction, you'll see that it is easy to arrange that $a_n=0$ for $n<0$. Then $\sum_{n=0}^{\infty} a_n z^n$ defines an analytic function inside the unit disc, which extends continuously to the boundary, but such that $\sum a_n$ is divergent.
Below, I give the details. I used Pinsky's book as my reference for the construction of du Bois - Reymond.
Define $S_M(z) = \sum_{r=1}^M \frac{z^r-z^{-r}}{r}$. Our function will be of the form
$$f(z) := \sum_{k=1}^{\infty} \frac{z^{N_k}}{k^2} S_{M_k}(z) \quad (1)$$
where $M_k$ and $N_k$ are sequences of positive integers chosen such that
$$0 < N_1 - M_1 < N_1 + M_1 < N_2 - M_2 < N_2 + M_2 < N_3 - M_3 < N_3+ M_3 < \cdots \quad (2)$$
and
$$\frac{\log M_k}{k^2} \to \infty \ \mbox{as} \ k \to \infty. \quad (3)$$
We will show below that the sum (1) is uniformly convergent in the closed unit disc. Hence, it defines a continuous function on the closed disc and an analytic function in the interior. Condition (2) forces that the polynomials $z^{N_k} s_{M_k}(z)$ have no overlapping terms, so the Taylor series of $f$ just looks like blocks of $-1/M_k$, $-1/(M_k-1)$, ..., $-1$, $0$, $1$, $1/2$, ..., $1/M_k$, separated by long blocks of zeroes. Define $a_n$ to be the coefficients of $f(z) = \sum_{n=0}^{\infty} a_n z^n$.
We now check that $\sum a_n$ is divergent. We have
$$\sum_{n=0}^{N_k} a_n = \sum_{j=1}^{k-1} \frac{1}{j^2} \left( \frac{-1}{M_j} + \cdots + \frac{-1}{1} + \frac{1}{1} + \cdots + \frac{1}{M_j} \right) - \frac{1}{k^2} \left( \frac{1}{M_k} + \frac{1}{M_k -1} + \cdots + \frac{1}{2} + \frac{1}{1} \right)$$
$$=0+0+\cdots + 0 - \frac{1}{k^2} \left( \frac{1}{M_k} + \frac{1}{M_k -1} + \cdots + \frac{1}{2} + \frac{1}{1} \right) \approx - \frac{\log M_k}{k^2}.$$
Using condition (3), this goes to $- \infty$. So there is a subsequence of partial sums of $\sum a_n$ which goes to $- \infty$ and $\sum a_n$ diverges.
We now must prove that (1) is uniformly convergent. We need
Lemma There is an absolute constant $C$ so that, for any real angle $\theta$, and any positive integer $M$, we have
$$\left| \sum_{r=1}^M \frac{\sin (r \theta)}{r} \right| \leq C.$$
Given the lemma, uniform convergence is easy. By the maximum modulus principle, $\left| z^{N_k} s_{M_k}(z) \right|$ is maximized on the boundary of the unit disc.
On that boundary, $|e^{i N_k \theta} s_{M_k}(e^{i \theta} )| = 2 \left| \sum_{r=1}^{M_k} \frac{\sin (r \theta)}{r} \right|$ and so, by the lemma, is bounded independently of $\theta$. The $\frac{1}{k^2}$ factor in front then forces uniform convergence.
We now prove the lemma. This is the only part I'm not adapting from Pinsky, because he treats this as obvious.
Proof of lemma: Since the sum of sines is clearly periodic modulo $2 \pi$, and is clearly odd, we may assume that $\theta \in (0,\pi)$. We break the sum at $r=K$, for a parameter $K$ to be chosen later. For the first part of the sum,
$$\left| \sum_{r=1}^{K} \frac{\sin (r \theta)}{r} \right| \leq \sum_{r=1}^{K} \frac{r \theta}{r} = K \theta.$$
For the second part of the sum, we start by noting
$$\sin \frac{\theta}{2} \cdot \sum_{r=K+1}^M \frac{\sin (r \theta)}{r} = $$
$$\frac{\cos ((K+1/2) \theta)}{K+1} + \sum_{r=K+2}^M \cos ((r-1/2) \theta)\left( \frac{1}{r-1} - \frac{1}{r} \right)+ \frac{\cos((M+1/2) \theta)}{M}$$
so
$$\left| \sum_{r=K+1}^M \frac{\sin (r \theta)}{r} \right| \leq \frac{1}{\sin (\theta/2)} \left( \frac{1}{K+1} + \sum_{r=K+2}^M \left( \frac{1}{r-1} - \frac{1}{r} \right)+ \frac{1}{M}\right) $$
$$ = \frac{2}{\sin (\theta/2) (K+1)} \leq \frac{2}{(K+1)(\theta/\pi)} = \frac{2 \pi}{(K+1)\theta}.$$
We have used the bound $\sin (\theta/2) \geq \theta/\pi$ for $\theta \in (0 , \pi)$, which is true because $\sin$ is concave.
In short,
$$\left| \sum_{r=1}^M \frac{\sin (r \theta)}{r} \right| \leq K \theta + \frac{2 \pi}{(K+1) \theta}.$$
Choose $K$ such that $K \theta$ is neither near $0$ nor $\infty$, and this quantity will be bounded, so we have proved the lemma. $\square$.
Best Answer
The function $f$ can be written as
$$ f(z)=g(z)+\sum_{k=1}^N\frac{a_k}{z-z_k} $$
for some $r>1$ and $|z|<r$. Here $g$ is holomorphic and $|z_k|\geq 1$ for $k\in\{1, \dotsc, N\}$. The boundedness of the coefficients follows directly from this decomposition (note that the radius of convergence of $g$ is at least $r>1$).