= Solution
Assume $\psi_+(c)=e^{-Ac}$ with $A\geq0$. Part (ii) gives
$$
\psi_-(c)=\frac{q}{a+A}e^{-Ac},\qquad a=q+\lambda.
$$
Insert this into (iii). If $A\ne a$, the convolution becomes
$$
e^{-Ac}=e^{-ac}+\frac{q^2}{(a+A)(a-A)}(e^{-Ac}-e^{-ac}).
$$
Equality for all $c>0$ forces $q^2=(a+A)(a-A)$, hence $A^2=a^2-q^2=\lambda(2q+\lambda)$. The case $A=a$ would instead give $e^{-ac}$ times a nonzero term linear in $c$, so is impossible. Thus
$$
\boxed{A=\sqrt{\lambda(2q+\lambda)},\qquad\psi_+(c)=e^{-c\sqrt{\lambda(2q+\lambda)}},\quad\psi_-(c)=\frac{q}{q+\lambda+A}e^{-Ac}.}
$$
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