= Solution
For every $i\in\mathbb N_0$, truncate $w$ before coordinate $i$:
$$
f_i(j)=\begin{cases}w(j),&j<i,\\0,&j\geq i.\end{cases}
$$
This $f_i$ lies in $F$. For $j<i$, $(w-f_i)(j)=0$. For $j\geq i$, $(w-f_i)(j)=w(j)=\pm2^j$, which is divisible by $2^i$. In particular the integer sequence
$$
g_i(j)=\begin{cases}0,&j<i,\\w(j)/2^i,&j\geq i\end{cases}
$$
belongs to $B$ and satisfies $w=f_i+2^ig_i$.
Apply the <module homomorphism> $\pi$. Since $f_i\in F\subseteq\ker\pi$,
$$
\pi(w)=2^i\pi(g_i)\qquad\text{for every }i\in\mathbb N_0.
$$
Fixing any coordinate $j$, the <integer> $\pi(w)(j)$ is therefore divisible by $2^i$ for every $i$. The only such <integer> is zero: if $a\ne0$, choose $i$ with $2^i>|a|$, which rules out $a=2^ib$ for any nonzero <integer> $b$. Thus every coordinate of $\pi(w)$ vanishes, contradicting part (b)(iii).
It follows that
$$
\boxed{B=\prod_{j\in\mathbb N_0}\mathbb Z\text{ is not a free }\mathbb Z\text{-module}}.
$$
The decisive property is <vanishing of infinitely divisible integer sequences>, namely $\bigcap_{i\geq0}2^iB=\{0\}$.
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