= Solution
The constant-coefficient equations and decay imply that only the modes $e^{iax}$ on the positive half-line and $e^{-ibx}$ on the negative half-line can occur; the opposite modes grow. <Continuity> and the nonzero <derivative> jump then imply nonzero amplitudes, so any common decay exponent satisfies $0<\gamma\leq\operatorname{Im}a<\operatorname{Im}b$. In particular $-b$ lies below the plus <holomorphic> half-plane, and $a$ lies outside the minus half-plane. This locates the <poles> needed for the transform calculation, without yet computing the amplitudes.
Rearrange the equation from part (i) as
$$
\frac{k+a}{k+b}Y_+-\frac1{(a+b)(k+b)}
=-\left(\frac{k-b}{k-a}Y_-+\frac1{(a+b)(k-a)}\right).
$$
The left side is <holomorphic> in the plus half-plane, and the right side in the minus half-plane. Their equality on the overlap glues them into one <entire function>. The assumed decay of $Y_\pm$, together with the rational factors tending to one and the subtracted fractions tending to zero, makes this <entire function> tend to zero at <infinity>. It is therefore identically zero by <Liouville's theorem>. This is the additive splitting step of the <Wiener-Hopf method>. Solving the two zero identities gives
$$
\boxed{Y_+(k)=\frac1{(a+b)(k+a)},\qquad Y_-(k)=-\frac1{(a+b)(k-b)}.}
$$
Their apparent cancellation formulas extend holomorphically wherever a cancelled factor vanishes; the only actual <poles> are at $-a$ and $b$.
With the Fourier convention in use, inversion has factor $e^{-ikx}$ and normalization $1/(2\pi)$. For $x>0$, close the <contour> in the <lower half-plane> clockwise, collecting the <pole> at $-a$. For $x<0$, close in the <upper half-plane> counterclockwise, collecting the <pole> at $b$. The <residue theorem> gives
$$
\boxed{y(x)=\begin{cases}\dfrac{e^{iax}}{i(a+b)},&x>0,\\[6pt]\dfrac{e^{-ibx}}{i(a+b)},&x<0.\end{cases}}
$$
Equivalently, each displayed transform follows immediately by integrating the corresponding decaying exponential on its half-line, so Fourier uniqueness gives the same inversion without a <contour> estimate. Both values at zero are $1/(i(a+b))$, and the <derivative> jump is $(a+b)/(a+b)=1$, directly checking the interface conditions.
The moduli are a nonzero constant times $e^{-\operatorname{Im}a\,x}$ for $x>0$ and $e^{-\operatorname{Im}b\,|x|}$ for $x<0$. Hence the largest common exponential decay rate is
$$
\boxed{\gamma_{\max}=\operatorname{Im}a.}
$$
Strictly, the stated big-$O$ bound is also true for every smaller positive $\gamma$; it does not specify a unique exponent unless the sharp rate is intended. The two transform convergence half-planes are exactly the sharper domains stated in part (ii), and the <poles> on their boundaries prevent extending either half-line <integral> domain any farther.
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