1. We use the given functional equations.

For $f$: $f(x+y)=f(x)+f(y)$ So $f$ is additive.

For $g$: $g(x+y)=g(x)g(y), \qquad g(x)>0$ So $g$ is exponential-type under addition.

2. First find $f\left(\frac14\right)$.

Given: $f\left(-\frac35\right)=12$ Using additivity, $f\left(-\frac35\right)=-f\left(\frac35\right)$ Hence, $f\left(\frac35\right)=-12$

Now, $f\left(\frac35\right)=f\left(3\cdot \frac15\right)=3f\left(\frac15\right)$ So, $3f\left(\frac15\right)=-12 \implies f\left(\frac15\right)=-4$

Also, $f\left(\frac14\right)=\frac54 f\left(\frac15\right)$ because for additive functions on rationals, $f(qx)=qf(x)$ for rational $q$.

Thus, $f\left(\frac14\right)=\frac54(-4)=-5$

A quicker way is: $\frac14=-\frac{5}{12}\left(-\frac35\right)$ so $f\left(\frac14\right)=-\frac{5}{12}\cdot 12=-5$

3. Now find $g(-2)$.

Given: $g\left(-\frac13\right)=2$ Using $g(nx)=g(x)^n$ for integer $n$, we get $g(-2)=g\left(6\cdot -\frac13\right)=\left(g\left(-\frac13\right)\right)^6=2^6=64$

4. Find $g(0)$.

Using the functional equation: $g(0)=g(0+0)=g(0)^2$ So, $g(0)=0 \text{ or } 1$ But $g(x)\in(0,\infty)$, so $g(0)>0$. Hence, $g(0)=1$

5. Evaluate the required expression: $\left(f\left(\frac14\right)+g(-2)-8\right)g(0)$ Substitute values: $=\left(-5+64-8\right)(1)$ $=51$

6. Comparison with stored correct answer.

Derived answer: $51$

Stored correct answer: $51$

Hence, the answer matches.