= Solution
Write $x\in Z(A\otimes_kB)$ uniquely as $x=\sum_j a_j\otimes b_j$ using part (i). For each $a\in A$, the equality $(a\otimes1)x=x(a\otimes1)$ becomes
$$
\sum_j(aa_j-a_ja)\otimes b_j=0.
$$
The coefficient uniqueness from part (i) gives $aa_j=a_ja$ for every $a$, so each $a_j$ belongs to the <center of an associative algebra> $Z(A)$. Consequently $x\in Z(A)\otimes_kB$.
Choose a <basis> $z_1,\ldots,z_t$ of the <vector space> $Z(A)$ and rewrite $x=\sum_{i=1}^t z_i\otimes d_i$. For $b\in B$, commuting with $1\otimes b$ gives
$$
\sum_i z_i\otimes(bd_i-d_ib)=0.
$$
Extend the <linearly independent> family $z_i$ to a <basis> of $A$ and apply its <dual basis> to the first factor. It follows that $bd_i=d_ib$ for every $b$, hence $d_i\in Z(B)$. This proves $Z(A\otimes_kB)\subseteq Z(A)\otimes_k Z(B)$.
Conversely, if $z\in Z(A)$ and $w\in Z(B)$, then
$$
(z\otimes w)(a\otimes b)=za\otimes wb=az\otimes bw=(a\otimes b)(z\otimes w).
$$
The <pure tensors> $a\otimes b$ span the <tensor product>, so $z\otimes w$ is central, as is every sum of such <pure tensors>. Therefore the <center of a tensor product of associative algebras> is
$$
\boxed{Z(A\otimes_kB)=Z(A)\otimes_k Z(B).}
$$
These <tensor products> are identified with their natural <vector subspaces> of $A\otimes_kB$; extending <bases> justifies these embeddings over the <field> $k$.
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