= Solution
There is a small closure issue in the printed set: the <unit disc> is open, so its half-disc is not relatively closed in $\mathbb H$. As written, the set is not a <compact H-hull>. We compute the intended <half-plane capacity> after taking its relative closure in $\mathbb H$.
First remove the closed unit half-disc $K$. Its <mapping-out function of a compact H-hull> is the <Joukowski map>
$$
g_K(z)=z+\frac1z,\qquad\operatorname{hcap}(K)=1.
$$
For $z=iy$ on the remaining portion of the vertical slit, $1<y\leq2$,
$$
g_K(iy)=i\left(y-\frac1y\right).
$$
Thus the image of the remaining slit is $(0,3i/2]$. By the <half-plane capacity of a vertical slit>, its capacity is $(3/2)^2/2=9/8$. The <half-plane-capacity composition rule> gives
$$
\boxed{\operatorname{hcap}(\overline A\cap\mathbb H)=1+\frac98=\frac{17}{8}.}
$$
As a direct check, composing with the slit map gives $g_A(z)=\sqrt{(z+z^{-1})^2+9/4}$, with the branch asymptotic to $z$. Its $z^{-1}$ coefficient is $17/8$.
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