= Solution
By the <monoidal coherence theorem>, calculations may suppress the canonical <associators> and <unitors>, restoring them uniquely afterwards. In this notation the two <Frobenius monoidal functor> identities read
$$
\psi_{A\otimes B,C}\varphi_{A,B\otimes C}=(\varphi_{A,B}\otimes1)(1\otimes\psi_{B,C}),
$$
$$
\psi_{A,B\otimes C}\varphi_{A\otimes B,C}=(1\otimes\varphi_{B,C})(\psi_{A,B}\otimes1).
$$
Write
$$
E=\psi_0F(e)\varphi_{X,Y},\qquad N=\psi_{Y,X}F(n)\varphi_0.
$$
We check both <snake identities> for this prospective <dual pair in a monoidal category>. For $FX$, expand and use the first Frobenius identity:
$$
\begin{aligned}
(E\otimes1)(1\otimes N)&=(\psi_0\otimes1)(Fe\otimes1)(\varphi_{X,Y}\otimes1)(1\otimes\psi_{Y,X})(1\otimes Fn)(1\otimes\varphi_0)\\
&=(\psi_0\otimes1)(Fe\otimes1)\psi_{X\otimes Y,X}\varphi_{X,Y\otimes X}(1\otimes Fn)(1\otimes\varphi_0)\\
&=(\psi_0\otimes1)\psi_{I,X}F(e\otimes1)F(1\otimes n)\varphi_{X,I}(1\otimes\varphi_0)\\
&=1_{FX}.
\end{aligned}
$$
The third line uses <naturality> of $\psi$ and $\varphi$. The last line uses $(e\otimes1)(1\otimes n)=1_X$, followed by the opmonoidal and monoidal unit axioms.
For $FY$, the second Frobenius identity gives the other calculation:
$$
\begin{aligned}
(1\otimes E)(N\otimes1)&=(1\otimes\psi_0)(1\otimes Fe)(1\otimes\varphi_{X,Y})(\psi_{Y,X}\otimes1)(Fn\otimes1)(\varphi_0\otimes1)\\
&=(1\otimes\psi_0)(1\otimes Fe)\psi_{Y,X\otimes Y}\varphi_{Y\otimes X,Y}(Fn\otimes1)(\varphi_0\otimes1)\\
&=(1\otimes\psi_0)\psi_{Y,I}F(1\otimes e)F(n\otimes1)\varphi_{I,Y}(\varphi_0\otimes1)\\
&=1_{FY}.
\end{aligned}
$$
Thus \b[both triangular identities hold], and
$$
\boxed{E:FX\otimes FY\to I,\qquad N:I\to FY\otimes FX}
$$
are the <evaluation morphism> and <coevaluation morphism> of a <dual pair in a monoidal category>. Notice that none of the comparison maps was assumed invertible.
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