= Solution
Let $\pi:F\to H$ send the free generators to the given presentation generators, and let $N=\ker\pi$, the <normal closure> of the relators $w_j$. The <Mihailova subgroup> here is exactly the <fiber product of groups>
$$
\boxed{L=\{(u,v)\in F\times F:\pi(u)=\pi(v)\}.}
$$
To prove this equality, each specified generator belongs to the displayed fiber product, so $L$ is contained in it. The diagonal generators produce every $(a,a)$. Conjugating $(1,w_j)^{\pm1}$ by a diagonal element gives $(1,a^{-1}w_j^{\pm1}a)$; products give all $(1,n)$ for $n\in N$. Conversely, if $\pi(u)=\pi(v)$, then $u^{-1}v\in N$ and
$$
(u,v)=(u,u)(1,u^{-1}v)\in L.
$$
In particular $(1,w)\in L$ if and only if $w=1$ in $H$. Thus membership in this finitely generated subgroup is undecidable, despite the soluble <word problem for a group> in the ambient <direct product of groups> $F\times F$.
For an explicit finite presentation, use $a_i$ and $b_i$ for the first and second copies of the free basis. The base has presentation $\langle a_1,\ldots,a_m,b_1,\ldots,b_m\mid[a_i,b_j]=1\ (1\leq i,j\leq m)\rangle$. The pairs $(x_i,x_i)$ and $(1,w_j(x))$ are $a_ib_i$ and $w_j(b)$. Consequently the requested <HNN extension> has the finite presentation
$$
\boxed{G=\left\langle a_1,\ldots,a_m,b_1,\ldots,b_m,t\ \middle|\
[a_i,b_j]=1,\ [t,a_ib_i]=1,\ [t,w_j(b)]=1\right\rangle.}
$$
Commutation with the displayed subgroup generators implies commutation with all their products and inverses, so these finitely many relations are exactly the specified centralizing extension. The associated subgroup is $L$ on both sides, and the associated isomorphism is the identity.
For an input word $w$ in the generators of $H$, form $p=w(b)\in P$ and the effectively constructed word
$$
W=t^{-1}ptp^{-1}.
$$
If $w=1$ in $H$, then $p=(1,w)\in L$ and $W=1$ by centralization. If $w\ne1$, then $p\notin L$; the displayed word contains stable letters and no <pinch in an HNN extension>, so <Britton's lemma> gives $W\ne1$. This is the <centralizing HNN extension detects subgroup membership> equivalence
$$
\boxed{W=1\text{ in }G\quad\Longleftrightarrow\quad w=1\text{ in }H.}
$$
A solution of the <word problem for a group> in $G$ would therefore solve that in $H$, a contradiction. \b[The finitely presented extension $G$ has insoluble word problem.]
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