= Solution
With total density one, the open <probabilities> for the individual gates are $m=S_{10}+S_{11}$ and $h=S_{01}+S_{11}$. Adding the relevant <master equations> cancels transitions of the other gate:
$$
\boxed{\dot m=\alpha(S_{00}+S_{01})-\beta(S_{10}+S_{11})
=\alpha(1-m)-\beta m,}
$$
$$
\boxed{\dot h=\gamma(S_{00}+S_{10})-\delta(S_{01}+S_{11})
=\gamma(1-h)-\delta h.}
$$
These marginal <ordinary differential equations> hold even for correlated initial gates. Their product gives $S_{11}$ when the independence assumption in part (c) holds. Solving the <first-order linear differential equations> from the all-closed initial state gives
$$
m(t)=\frac{\alpha}{\alpha+\beta}\bigl(1-e^{-(\alpha+\beta)t}\bigr),\qquad
h(t)=\frac{\gamma}{\gamma+\delta}\bigl(1-e^{-(\gamma+\delta)t}\bigr).
$$
Put $\lambda=\alpha+\beta=\gamma+\delta>0$. Then
$$
S_{11}(t)=S_{11,\infty}(1-e^{-\lambda t})^2,\qquad
S_{11,\infty}=\frac{\alpha\gamma}{\lambda^2}.
$$
For positive opening rates this increases strictly from zero, so its half-<equilibrium> time is uniquely determined by $(1-e^{-\lambda t})^2=1/2$:
$$
\boxed{t_{1/2}=\frac{-\log(1-2^{-1/2})}{\lambda}
=\frac{\log(2+\sqrt2)}{\lambda}.}
$$
If an opening rate is zero, the all-open <equilibrium> density is zero and a nontrivial half-rise time is undefined.
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