= Solution
In an <Einstein-de Sitter universe>, $H(a)=H_0a^{-3/2}$, so direct integration of $dt=da/(aH)$ gives $t(a)=2a^{3/2}/(3H_0)$. At $z=8$, $a=1/9$ and $a^{3/2}=1/27$. The fraction of the present <age of an FLRW universe> and the available formation interval are therefore
$$
\boxed{\frac{t(8)}{t_0}=\frac1{27}\simeq0.0370,
\qquad t(8)=\frac{2}{81H_0}
=2.47\times10^8h^{-1}\,\mathrm{yr}.}
$$
This is only about $3.7\%$ of the present age. The available interval is $t(8)$ itself, not $t_0-t(8)$, which would instead be the lookback time.
For $\Omega_{m,0}=0.3$ and $\Omega_{\Lambda,0}=0.7$, at the same <Hubble constant>, the exact integral gives
$$
t(8)=\frac{2}{3H_0\sqrt{0.7}}
\operatorname{arsinh}\!\left(\frac{\sqrt{0.7/0.3}}{27}\right)
\simeq0.04506H_0^{-1}
=4.51\times10^8h^{-1}\,\mathrm{yr}.
$$
Thus \b[the formation interval is longer, by about a factor of $1.83$]. There is also a direct comparison: for $0<a<1$, $0.3a^{-3}+0.7<a^{-3}$, so $H(a)$ is smaller than its <Einstein-de Sitter universe> value at fixed $a$, and integrating $da/(aH)$ gives a greater age. At $z=8$ the leading matter-era result is $t\simeq2/[3H_0\sqrt{0.3}(1+z)^{3/2}]$; the smaller matter density, with the same $H_0$, explains nearly all the increase.
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