$$ \begin{aligned} & \mathrm{F}_1=\mathrm{mg} \sin 45^{\circ}+\mathrm{f}=\mathrm{mg} \sin 45^{\circ}+\mu \mathrm{N} \\\\ & \mathrm{F}_1=\frac{\mathrm{mg}}{\sqrt{2}}+\mu \mathrm{mg} \cos 45^{\circ} \\\\ & \mathrm{F}_1=\frac{\mathrm{mg}}{\sqrt{2}}(1+\mu) \end{aligned} $$

$$ \begin{aligned} & F_2=m g \sin 45^{\circ}-f=m g \sin 45^{\circ}-\mu N \\\\ & =\frac{m g}{\sqrt{2}}(1-\mu) \end{aligned} $$

$$ \begin{aligned} & \mathrm{F}_1=2 \mathrm{~F}_2 \\\\ & \frac{\mathrm{mg}}{\sqrt{2}}(1+\mu)=2 \frac{\mathrm{mg}}{\sqrt{2}}(1-\mu) \\\\ & 1+\mu=2-2 \mu \\\\ & \mu=1 / 3=0.33 \end{aligned} $$