= Solution
A <transitive set> $x$ satisfies $z\in y\in x\Rightarrow z\in x$, equivalently $\bigcup x\subseteq x$. If $x$ is transitive and $z\in y\in\bigcup x$, choose $w\in x$ with $y\in w$. Transitivity gives $y\in x$, so $z\in\bigcup x$. Thus $\bigcup x$ is transitive. If $z\in y\in\mathcal P(x)$, then $z\in x$ and transitivity implies $z\subseteq x$, whence $z\in\mathcal P(x)$. Thus the <power set> is transitive as well.
The converse for the union is false: take $x=\{\{\varnothing\}\}$. Then $\bigcup x=\{\varnothing\}$ is transitive but $x$ is not. The power-set converse is true: if $z\in y\in x$, then $y\in\{y\}\in\mathcal P(x)$ implies $y\in\mathcal P(x)$, so $z\in x$.
Define $x_0=x$, $x_{n+1}=\bigcup x_n$ and $\operatorname{TC}(x)=\bigcup_{n<\omega}x_n$. This is a set by <replacement> and <union>, contains $x$ as a subset, and is transitive because taking a member advances one stage. Induction shows it is contained in every transitive set containing $x$, proving the <transitive closure> property.
If the <rank of a set> $x$ is $\alpha$, then
$$
\boxed{\operatorname{rank}\mathcal P(x)=\alpha+1,\qquad\operatorname{rank}\operatorname{TC}(x)=\alpha.}
$$
Every subset of $x$ has rank at most $\alpha$, and $x$ itself is an element of $\mathcal P(x)$, proving the first formula. The second is <transitive closure preserves set-theoretic rank>. Here the convention is $x\subseteq\operatorname{TC}(x)$, not $x\in\operatorname{TC}(x)$.
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