= Solution
For the final low steady value $Z_s=\beta_{\rm low}/\alpha$, the first crossing of $Z_s/2$ occurs during high production if $F(1-e^{-\alpha d})\ge1/2$. Solving $F(1-e^{-\alpha t_{1/2}})=1/2$ then gives, for either motif,
$$
\boxed{t_{1/2}=\frac1\alpha\ln\frac{2F}{2F-1}
\simeq\frac1{2\alpha F}\quad(F\gg1).}
$$
This <half-final response of a graded incoherent feed-forward loop> is a first-crossing measure. A simple output produced at the final low rate from zero would instead take $(\ln2)/\alpha$. This comparison explains the acceleration of the first half-final response, not the time needed to settle after a potentially large overshoot.
The condition matters because the source does not fix $K_{yz}/Y_s$. Define $\delta=1-e^{-\alpha d}$, equal to $h$ for <I1 incoherent feed-forward loop> and $1-h$ for <I4 incoherent feed-forward loop>. If the high phase ends before half-final output is reached, solve the second branch of part (a) instead:
$$
\boxed{t_{1/2}=\begin{cases}
\alpha^{-1}\ln[2F/(2F-1)],&F\delta\ge1/2,\\
d+\alpha^{-1}\ln[2(1-F\delta)],&F\delta<1/2.
\end{cases}}
$$
The expressions agree at the boundary. The $F$-only answer is therefore valid for both motifs when $F\min(h,1-h)\ge1/2$, as holds for large F and a fixed interior threshold. Merely saying $F\gg1$ does not cover a threshold that approaches zero or one even faster. If the Y threshold is never crossed, the assumed final low state itself changes and one must reconsider the <steady state> before using this response-time definition.
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