The first set is separable but not complete.

A dense countable subset is given by finite, zero-sum rational sequences. Let $\epsilon > 0$, and $a \in A$. Since $\sum_{k=1}^\infty a_k = 0$, then there exists $n$ such that $\lvert a_k \rvert < \epsilon/2$ for $k > n$. Now let $q_1, \dots, q_n$ be rational numbers such that $\lvert q_k - a_k < \epsilon$. Let $Q = \sum_{k=1}^n q_k$, and let $m$ such that $Q/m < \epsilon/2$. Then the sequence defined by

- $b_k = q_k$ for $k \leq n$
- $b_k = -Q/m$ for $n < k \leq m+n$
- $b_k = 0$ for $k > m+n$

is a finite zero-sum rational sequence, and $d(a, b) < \epsilon$ by construction.

To show that it is not complete, let $\{ a^{(n)} \mid n \in \mathbb{N} \}$ be a sequence of sequences defined by

- $a^{(n)}_0 = 1$ for every $n$
- $a^{(n)}_k = 0$ for $0 < k < n$ and $k \geq 2n$
- $a^{(n)}_k = -1/n$ for $n \leq k < 2n$.

It is immediate that $a^{(n)} \in A$ for every $n$, and it is also clear that these are Cauchy sequences: for every $\epsilon > 0$, let $n$ be such that $2/n < \epsilon$. Then for every $m, m' > n$, indeed \[ d_\infty(a^{(m)}, a^{(m')}) < \frac{1}{m} + \frac{1}{m'} < \frac{1}{n} + \frac{1}{n} < \epsilon \] so it is a Cauchy sequence. However, $\lim_{n \rightarrow \infty} a^{(n)}_0 = 1$ and $\lim_{n \rightarrow \infty} a^{(n)}_k = 0$ for $k > 0$, so the limit is the sequence $1, 0, 0, \dots$ which is obviously not in $A$.

The second set is also separable but not complete. Let $B$ be the subset defined by the same conditions, but $c_k \in \mathbb{Q}$. Then $B$ is obviously countable. Let $f \in A, f(x) = \sum_{k=1}^n c_k x^k$, and let $\epsilon > 0$. Let $q_1, \dots, q_n$ be between $0$ and $1$ and such that $\lvert q_k - c_k \rvert < \epsilon/2n$, and let $g(x) = \sum_{k=1}^n q_k x^k \in B$. Then \[ d_\infty(f, g) = \max_x \left\lvert \sum_{k=1}^n (c_k - q_k) x^k \right\rvert \leq \sum_{k=1}^n \lvert c_k - q_k \rvert \leq n \cdot \frac{\epsilon}{n} = \epsilon \] so $B$ is dense and so $A$ is separable.

Let \[ f_k(x) = \sum_{i=1}^k \frac{1}{i!} x^i. \] These are all polynomials in $A$, and it is well known that they converge uniformly to $e^x - 1$, so in particular they form a Cauchy sequence in $A$, but the limit is not in $A$, and so $A$ is not complete.