= Solution
The <bounding number> $\mathfrak b$ is the least size of an unbounded family in $(\omega^\omega,\le^*)$. The <almost disjointness number> $\mathfrak a$ is the least size of an infinite <maximal almost disjoint family on omega> of infinite <subsets> of $\omega$. Requiring the family to be infinite excludes trivial finite maximal partitions.
A countable family $\{f_i:i<\omega\}$ is bounded by $g(n)=1+\max_{i\le n}f_i(n)$. Thus $\aleph_1\le\mathfrak b$.
Suppose an infinite <almost disjoint family on omega> $\mathcal A$ has size less than $\mathfrak b$. Choose distinct members $A_n$ and put $C_n=A_n\setminus\bigcup_{i<n}A_i$. These are infinite and pairwise disjoint. For each $B\in\mathcal A$, define $h_B(n)=\max((B\cap C_n)\cup\{0\})$ whenever the intersection is finite. If $B=A_n$ at the exceptional index $n$, <set> $h_B(n)=0$. These are the only possible infinite intersections. A single $g$ eventually dominates every $h_B$, because there are fewer than $\mathfrak b$ of them. Pick $x_n\in C_n$ with $x_n>g(n)$ and put $X=\{x_n:n<\omega\}$.
For $B$ not among the selected $A_n$, only finitely many $x_n$ can lie in $B$. The same is true for $B=A_j$, with its single exceptional index $j$ ignored. Thus $X$ is infinite and almost disjoint from every member of $\mathcal A$, so the family was not maximal. This <bounding-to-almost-disjointness inequality> proves $\mathfrak b\le\mathfrak a$.
Finally start with an infinite pairwise disjoint family and use <Zorn's lemma> to extend it to a <maximal almost disjoint family on omega>. It is a family of <subsets> of $\omega$, so has <cardinality> at most $2^{\aleph_0}$. Therefore
$$
\boxed{\aleph_1\le\mathfrak b\le\mathfrak a\le2^{\aleph_0}.}
$$
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