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. $\square$.
The thought process should probably begin with: let's look at some function with a simple pole on the unit circle, and with a power series that I can find. This should remind of the example
$$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$
Here the coefficients are all equal to $1$, so they are bounded. More generally, we can place a simple pole at any point $a$ on the unit circle, and give it the residue of $b$, like this:
$$\frac{b}{z-a}=\frac{-\bar a b}{1-\bar az} = -\bar a b \sum_{n=0}^\infty \bar a^n z^n$$
Again, the coefficients are bounded. And if we form a finite sum of such functions, the coefficients will be bounded still...
This sparks an idea (spoilered):
Subtract such a sum from given $f$ to cancel out the poles. The resulting function will have a radius of convergence greater than $1$, hence its coefficients will tend to $0$.
Best Answer
It seems to me that this is a particular case of an old Theorem from Cantor (1870), called Cantor's uniqueness theorem. The theorem says that if, for every real $x$, $$\lim_{N \rightarrow \infty} \sum_{n=-N}^N c_n e^{inx}=0,$$ then all the complex numbers $c_n$'s are zero.
You can google "Uniqueness of Representation by Trigonometric Series" for more information. See e.g. this document for a proof and some history of the result.