= Solution
At weak freeze-out, <photons>, electrons, positrons and all four <neutrino> flavors are relativistic and approximately at the same temperature. The <effective number of relativistic energy degrees of freedom> changes from
$$
g_*^{(3)}=2+\frac78(4+6)=\frac{43}{4},\qquad
\boxed{g_*^{(4)}=2+\frac78(4+8)=\frac{50}{4}.}
$$
The expansion rate at fixed temperature is larger by $\sqrt{50/43}\simeq1.078$. Equating the supplied weak interaction time with the supplied age gives $T_f^3=\sqrt{g_*}$ when $T_f$ is measured in MeV. Therefore
$$
\boxed{\frac{T_f^{(4)}}{T_f^{(3)}}=\left(\frac{50}{43}\right)^{1/6}
\simeq1.0255.}
$$
Earlier <cosmological weak freeze-out> at a higher temperature leaves more neutrons. With negligible chemical potentials, the <neutron-to-proton ratio> in equilibrium is $r_f=\exp(-Q/T_f)$, where $Q=m_n-m_p\simeq1.293\,\mathrm{MeV}$. If almost all surviving neutrons enter <helium-4>, the <primordial helium mass fraction> before neutron decay is $Y_f=2r_f/(1+r_f)$. Literal use of the given rate normalizations yields
$$
\begin{array}{c|cc}
&3\text{ flavors}&4\text{ flavors}\\\hline
T_f\ (\mathrm{MeV})&1.486&1.523\\
r_f&0.419&0.428\\
Y_f&0.590&0.599
\end{array}
\qquad
\boxed{\Delta Y_f\simeq0.0090>0.}
$$
Equivalently, the leading response is $\delta Y_f/Y_f=[Q/(T_f(1+r_f))]\delta T_f/T_f$. This is a roughly one-percentage-point absolute increase in the helium fraction, rather than a one-percent relative increase.
The <deuterium bottleneck> delays nuclear burning, allowing <free-neutron decay after freeze-out>. If its end temperature is $T_N$, and $\tau_n$ is the neutron mean life, a sharper instantaneous-freeze-out estimate conserves total baryon number and gives
$$
Y_p\simeq\frac{2}{1+e^{Q/T_f}}
\exp\!\left[-\frac{t_N-t_f}{\tau_n}\right].
$$
This formula evolves the neutron fraction, not just the <neutron-to-proton ratio>, because each neutron decay also creates a proton. After electron annihilation, let $r_T=(4/11)^{1/3}$. The radiation counts relevant to the age at fixed $T_N$ are
$$
g_{*,N}^{(3)}=2+\frac78(6)r_T^4\simeq3.36,\qquad
g_{*,N}^{(4)}=g_{*,N}^{(3)}+\frac74r_T^4\simeq3.82.
$$
Consequently $t_N^{(4)}/t_N^{(3)}\simeq\sqrt{3.36/3.82}\simeq0.939$. Less time is available for neutron decay, adding to the helium increase. As an explicit illustrative completion, $T_N=0.07\,\mathrm{MeV}$ and $\tau_n\simeq880\,\mathrm{s}$ give an increase of approximately $0.012$ in this toy model. The precise helium fraction requires the nuclear-burning temperature, baryon abundance and properly normalized weak rates; the question supplies only approximate time laws. In particular, the large no-decay fractions in the table must not be interpreted as precise predictions for the observed primordial abundance. The robust effect is \b[extra relativistic energy raises the primordial helium fraction], with an absolute change of order $10^{-2}$ for one thermally populated flavor.
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