= Solution
Let $1$ be a <terminal object>. The <pullback> of the unique arrows $A\to1\leftarrow B$ is a binary <categorical product>: maps $X\to A$ and $X\to B$ automatically have equal composites into $1$, so the <pullback> <universal property> is exactly the <categorical product> <universal property>.
For parallel arrows $f,g:A\rightrightarrows B$, form $B\times B$ and take the <pullback> of $\langle f,g\rangle:A\to B\times B$ along the diagonal $\delta:B\to B\times B$. Write the resulting projections as $e:E\to A$ and $b:E\to B$. The <pullback> equation gives
$$
fe=b=ge.
$$
If $u:X\to A$ satisfies $fu=gu$, the pair $(u,fu)$ is a <pullback> cone, producing a unique $v:X\to E$ with $ev=u$ and $bv=fu$. The second condition is forced by the first, since $b=fe$. Thus $e$ is the <equaliser> of $f,g$.
The <terminal object> and iterated binary <categorical products> give all finite <categorical products>, including the nullary one. Part (a) therefore proves \b[<pullbacks> and a <terminal object> give all finite <categorical limits>].
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