= Solution
Let $x$ denote the generating object, the image of the unique object of the <terminal category>. The <free braided monoidal category on one object> can be described syntactically. Its objects are all fully parenthesized expressions built from $x$, the <monoidal unit object> $I$, and a binary <monoidal tensor product>. Thus $x\otimes(I\otimes x)$ and $(x\otimes I)\otimes x$ are distinct objects, though canonically isomorphic. Its <morphisms> are generated by the <associators>, <unitors>, <braidings> and their inverses, closed under composition and tensor product, subject to the pentagon, triangle, <naturality> and hexagon axioms. No symmetry relation is imposed on the <braiding>. The embedding of the <terminal category> selects $x$.
The <braid category> $\mathbf B$ has objects the nonnegative integers, with
$$
\mathbf B(m,n)=\begin{cases}B_n,&m=n,\\\varnothing,&m\ne n,\end{cases}
$$
where $B_n$ is the <braid group> on $n$ strands and $B_0,B_1$ are trivial. Composition is stacking braids, with the first morphism followed by the second. The <monoidal tensor product> is addition on objects and side-by-side juxtaposition of braids; its <monoidal unit object> is $0$. The <associators> and <unitors> are identities, so it is a <strict monoidal category>.
Choose the positive crossing convention once and for all. Its <braiding> $c_{m,n}:m+n\to n+m$ is the block braid moving the first $m$ strands over the next $n$ strands while retaining the order within each block. In particular $c_{1,1}=\sigma_1$. The usual <braid group> relations express the <naturality> and hexagon laws for these block braids. The generating <functor> $\mathbf1\to\mathbf B$ selects $1$.
\b[The distinction is parenthesized tensor expressions versus strand counts:] the first construction keeps the structural isomorphisms visible, while $\mathbf B$ has strict tensor arithmetic. These descriptions give the two requested free constructions without needing to establish their <universal properties>.
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