$\text{Let}f:R\to R\text{be a function such that }f(x+y)=f\left(x\right)+f(y) \text{for all}\mathrm{ x},y\in R,\text{and}g:R\to (0,\infty )$

$\text{be a function such that}g(x+y)=g\left(x\right)g\left(y\right)\text{for all}x,y\in R.\text{If}\mathrm{ f}\left(-\frac{3}{5}\right)=12\text{ and}\mathrm{ g}\left(-\frac{1}{3}\right)=2,$

$\text{then the value of }f\left(\frac{1}{4}\right)+g(-2)-8g\left(0\right)\text{is}\text{\_\_\_\_\_}$

The correct answer is 51.

Step 1: Determine the function $f\left(x\right)$

We are given that $f(x+y)=f\left(x\right)+f\left(y\right)$ for all real numbers $x,y$.

This is Cauchy's additive functional equation, which implies that $f\left(x\right)=kx$ for some constant $k$.

We are given:

$f(-\frac{3}{5})=12$

Substituting $x=-\frac{3}{5}$ into $f\left(x\right)=kx$:

$k(-\frac{3}{5})=12$

$k=12\times (-\frac{5}{3})=-20$

Therefore, the function is $f\left(x\right)=-20x$.

Now, we can find $f\left(\frac{1}{4}\right)$:

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

Step 2: Determine the values of $g\left(0\right)$ and $g(-2)$

We are given that $g(x+y)=g\left(x\right)g\left(y\right)$ for all $x,y\in R$ and $g:R\to (0,\infty )$.

To find $g\left(0\right)$, substitute $x=0$ and $y=0$:

$g(0+0)=g\left(0\right)g\left(0\right)\Rightarrow g\left(0\right)={\left(g\left(0\right)\right)}^2$

Since $g\left(x\right)>0$ for all $x$, we can divide by $g\left(0\right)$:

$g\left(0\right)=1$

To find $g(-2)$, note that:

$-2=6\times (-\frac{1}{3})$

Using the multiplicative property $g\left(nx\right)={\left(g\left(x\right)\right)}^n$:

$g(-2)=g{\left(-\frac{1}{3}\right)}^6$

Given $g(-\frac{1}{3})=2$, we get:

$g(-2)=2^6=64$

Step 3: Calculate the final value

We need to find the value of:

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

Substitute the evaluated values:

$=-5+64-8\left(1\right)$

$=59-8=51$