= Solution
The inverse <Bogoliubov transformation> gives $c=ua-vb^\dagger$ and $d=va^\dagger+ub$. Because $(a^\dagger b^\dagger)^2=0$ by the <canonical anticommutation relations>, the exponential truncates:
$$
Ae^{Ba^\dagger b^\dagger}|0\rangle
=A\bigl(|0\rangle+Ba^\dagger b^\dagger|0\rangle\bigr).
$$
The <fermionic signs> are fixed by
$$
a\,a^\dagger b^\dagger|0\rangle=b^\dagger|0\rangle,
\qquad
b\,a^\dagger b^\dagger|0\rangle=-a^\dagger|0\rangle.
$$
It follows that
$$
c|\tilde0\rangle=A(uB-v)b^\dagger|0\rangle,
\qquad
d|\tilde0\rangle=A(v-uB)a^\dagger|0\rangle.
$$
Both annihilation conditions hold precisely when $B=v/u$. Normalizing the <paired fermionic vacuum>, using the <orthogonality> of its two <Fock states>, gives $A^2(1+B^2)=1$. Choose the irrelevant overall phase so that $A>0$. The result is
$$
\boxed{A=u=\sqrt{\frac{E+\varepsilon}{2E}},\qquad
B=\frac vu=\frac{\Delta}{E+\varepsilon}
=\frac{E-\varepsilon}{\Delta},\qquad
E=\sqrt{\varepsilon^2+\Delta^2}.}
$$
Equivalently,
$$
\boxed{|\tilde0\rangle=u|0\rangle+va^\dagger b^\dagger|0\rangle.}
$$
Its positive pair coefficient follows from the specified order $ab+b^\dagger a^\dagger$ in the <Hamiltonian operator>. The state has even <fermion parity> and is a <quantum superposition> of zero and two original particles. In the limit $\Delta\to0^+$, $A\to1$ and $B\to0$, so it reduces to the original <vacuum state>, as expected for $\varepsilon>0$.
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