= Solution
Initially $a=1/2$ and $T=T_+/T_0>1$, so $\varepsilon T_t=1-T^4<0$. On the fast time $s=t/\varepsilon$, <albedo> changes by only $O(\varepsilon)$ and
$$
T_s=1-T^4,\qquad T(0)=T_+/T_0.
$$
Thus <temperature> rapidly falls towards $1$. At the same time $a_t=a^*(T)-1/2<0$ while $T>1$, which seeds selection of the warm branch. Afterwards the approximate <slow manifold> is $T^4=2(1-a)$: decreasing <albedo> makes <temperature> rise slowly back towards $T_+/T_0$. The preceding figure separates these two time scales; a single time axis would compress the initial fall into a nearly vertical segment.
In the sharp-transition regime, after selection of the warm branch $a^*(T)\simeq1/4$. Solving the linear <ordinary differential equation> with the leading matched value $a(0^+)=1/2$ gives
$$
\boxed{a(t)\simeq\frac14+\frac14e^{-t},\qquad T(t)\simeq\left(\frac{3-e^{-t}}2\right)^{1/4}.}
$$
These are <outer solutions>: their initial <temperature> is $1$, so the fast cooling layer is required to satisfy the actual initial <temperature>.
There is a useful qualification to the two-limit argument. For fixed finite $\mu>2$, put $x=1-2a$ on the <slow manifold>. Then $x_t=-x+\tfrac12\tanh(\mu x)$, and $x=0$ is an unstable <equilibrium> with growth rate $\mu/2-1$. The small positive $O(\varepsilon)$ seed from the fast layer can therefore produce a delay of order $\log(1/\varepsilon)/(\mu/2-1)$. The simple exponential approximation assumes that the sharp <albedo> transition is sufficiently fast to make this delay negligible on the displayed slow scale. It is not a uniform approximation for every fixed $\mu>2$ as $\varepsilon\to0$.
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