F_{2}=m g \sin 45^{\circ}-f
here f=\mu N, N=m g \cos 45^{\circ}
F_{2}=\frac{m g}{\sqrt{2}}-\mu m g \cos 45^{\circ}
F_{2}=\frac{m g}{\sqrt{2}}(1-\mu)
F_{1}=2 F_{2} (Given)
\frac{m g}{\sqrt{2}}(1+\mu)=2 \frac{m g}{\sqrt{2}}(1-\mu)

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