Let $f: R \to R$ be a function such that $f(x+y)=f(x)+f(y)$ for all x, y in R and $g: R \to (0,$ infinity) be a function such that $g(x+y)=g(x)g(y)$ for all x, y in R. If $f(-\frac{3}{5})=12$ and $g(-\frac{1}{3})=2,$ then the value of $(f(\frac{1}{4})+g(-2)-8)g(0)$ is _______.

Let f: R → R be a function such that f(x+y)=f(x)+f(y) for all x, y ∈ R and g: R → (0, ∞) be a function such that g(x+y)=g(x) g(y) for all x, y ∈ R. If f((-3)/5)=12 and g((-1)/3)=2, then the value of (f(1/4)+g(-2)-8) g(0) is _______.

$\text{Given: } f(x+y)=f(x)+f(y),\ \forall\, x,y \in \mathbb{R}$

$\text{Given: } g(x+y)=g(x)\,g(y),\ \ g(x) > 0$

$f\!\left(\frac{-3}{5}\right)=12,\qquad g\!\left(\frac{-1}{3}\right)=2$

$\text{To find: } \left(f\!\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)$

$\text{Plan: for rational } q,\ \ f(q)=q\,f(1)$

$g(q)=g(1)^{q}$

$\text{Find } f(1),\,g(1) \text{ from the two given values, then substitute.}$

$f(0)=f(0)+f(0)\ \Rightarrow\ f(0)=0$

$0=f(0)=f(x)+f(-x)\ \Rightarrow\ f(-x)=-f(x)$

$f(mx)=m\,f(x)\quad(m\in\mathbb{Z})$

$n\,f\!\left(\frac{1}{n}\right)=f(1)\ \Rightarrow\ f\!\left(\frac{1}{n}\right)=\frac{f(1)}{n}$

$\therefore\ f(q)=q\,f(1)\quad(q\in\mathbb{Q})$

$f\!\left(\frac{-3}{5}\right)=\frac{-3}{5}\,f(1)=12$

$\Rightarrow\ f(1)=12\times\frac{-5}{3}=-20$

$\Rightarrow\ f\!\left(\frac{1}{4}\right)=\frac{1}{4}(-20)=-5$

$g(0)=g(0)^2,\ \ g(0)>0\ \Rightarrow\ g(0)=1$

$1=g(0)=g(x)\,g(-x)\ \Rightarrow\ g(-x)=\frac{1}{g(x)}$

$g\!\left(\frac{-1}{3}\right)=g(1)^{-1/3}=2$

$\Rightarrow\ g(1)=2^{-3}=\frac{1}{8}$

$g(-2)=g(1)^{-2}=\left(\frac{1}{8}\right)^{-2}=64$

$\left(f\!\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)$

$=(-5+64-8)\times 1$

$=51$