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(C) Given $f(x+y)=f(x)+f(y)$. This is Cauchy's functional equation,which implies $f(x)=ax$ for all $x \in Q$.Given $f\left(\frac{-3}{5}\right)=12$,we

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$51$

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(C) Given $f(x+y)=f(x)+f(y)$. This is Cauchy's functional equation,which implies $f(x)=ax$ for all $x \in Q$.Given $f\left(\frac{-3}{5}\right)=12$,we have $a\left(\frac{-3}{5}\right)=12$,so $a=-20$. Thus,$f(x)=-20x$.Then $f\left(\frac{1}{4}\right)=-20\left(\frac{1}{4}\right)=-5$.Given $g(x+y)=g(x)g(y)$,this implies $g(x)=b^x$ for some $b>0$.Given $g\left(\frac{-1}{3}\right)=2$,we have $b^{-1/3}=2$,which means $b^{-1}=2^3=8$,so $b=\frac{1}{8}$.Thus,$g(x)=\left(\frac{1}{8}\right)^x=8^{-x}=2^{-3x}$.Then $g(-2)=2^{-3(-2)}=2^6=64$.Also,$g(0)=b^0=1$.Substituting these values into the expression: $\left(f\left(\frac{1}{4}\right)+g(-2)-8\right)g(0) = (-5+64-8)(1) = 51$.

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