Let $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x,y\in <!--\; \in \; -->\mathbb{R}$, and $g:\mathbb{R}\to <!--\; \to \; -->(0,\mathrm{\infty <!--\; \infty \; -->})$ be a function such that $g(x+y)=g(x)g(y)$ for all $x,y\in <!--\; \in \; -->\mathbb{R}$. If $f\left(\frac{-<!--\; -\; -->3}{5}\right)=12$ and $g\left(\frac{-<!--\; -\; -->1}{3}\right)=2$, then the value of  $(f\left(\frac{1}{4}\right)+g(-<!--\; -\; -->2)-<!--\; -\; -->8)g(0)$ is ______.

The correct answer is 51.

We are given two functional equations:
1) $f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x,y\in ℝ$, and
2) $g(x+y)=g\left(x\right)g\left(y\right)$ for all $x,y\in ℝ$, with $g:ℝ\to (0,\infty )$.

Let us first analyze the function $f\left(x\right)$. The Cauchy functional equation $f(x+y)=f\left(x\right)+f\left(y\right)$ has the general solution $f\left(x\right)=kx$ for rational numbers, and for all real numbers under standard continuity or boundedness assumptions.
We are given:
$f(-\frac{3}{5})=12$
Since $f\left(x\right)=kx$, we have:
$k(-\frac{3}{5})=12\Rightarrow k=12\times (-\frac{5}{3})=-20$
Thus, the function is:
$f\left(x\right)=-20x$

Now we calculate $f\left(\frac{1}{4}\right)$:
$f\left(\frac{1}{4}\right)=-20\times \frac{1}{4}=-5$

Next, let us analyze the function $g\left(x\right)$. The functional equation is $g(x+y)=g\left(x\right)g\left(y\right)$ with $g\left(x\right)>0$.
This is the exponential functional equation, which has the general form $g\left(x\right)=a^x$ for some constant $a>0$.
We are given:
$g(-\frac{1}{3})=2$
Substituting $g\left(x\right)=a^x$ gives:
$a^{-1/3}=2$
To find $g(-2)$, we compute:
$g(-2)=a^{-2}={\left(a^{-1/3}\right)}^6=2^6=64$

We also need the value of $g\left(0\right)$:
Since $g(x+y)=g\left(x\right)g\left(y\right)$ and $g\left(x\right)>0$ for all $x$:
$g\left(0\right)=g(0+0)=g\left(0\right)g\left(0\right)\Rightarrow g\left(0\right)\left(g\right(0)-1)=0$
Since $g\left(0\right)>0$, we have:
$g\left(0\right)=1$

Now, let us evaluate the expression:
$\left(f\right(\frac{1}{4})+g(-2)-8)g\left(0\right)$
Substituting the calculated values:
$(-5+64-8)\times 1=51\times 1=51$