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

The function $$f : \mathbb{R} \to \mathbb{R}$$ satisfies the additive rule
$$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}.$$ For such a function:

• $$f(0)=f(0+0)=f(0)+f(0)\;\Rightarrow\;f(0)=0$$.
• For any integer $$n$$, $$f(nx)=nf(x)$$ (apply the rule $$n$$ times).
• For any rational $$r=\dfrac{p}{q}$$, $$f(rx)=rf(x)$$ (use the integer result on $$q x$$ and divide).

Given $$f\!\left(-\dfrac35\right)=12$$, we first find $$f\!\left(\dfrac35\right)$$:
$$f\!\left(\dfrac35\right)=-f\!\left(-\dfrac35\right)=-12.$$ Now choose $$x=\dfrac35$$ and a rational multiplier $$r=\dfrac{1/4}{3/5}=\dfrac{5}{12}$$ so that $$r\,x=\dfrac14$$. Using the rational-multiple rule,

$$f\!\left(\dfrac14\right)=r\,f(x)=\dfrac{5}{12}\,(-12)=-5.$$

The function $$g : \mathbb{R} \to (0,\infty)$$ satisfies the multiplicative rule
$$g(x+y)=g(x)\,g(y)\quad\forall\,x,y\in\mathbb{R}.$$ For such a function:

• $$g(0)=g(0+0)=g(0)^2\;,\;g(0)\gt 0\;\Rightarrow\;g(0)=1.$$
• For any integer $$n$$, $$g(nx)=g(x)^n$$ (apply the rule $$n$$ times).
• Consequently, for rational $$r=\dfrac{p}{q}$$, $$g(rx)=g(x)^r$$ (use the integer result and take roots).

Given $$g\!\left(-\dfrac13\right)=2$$, we obtain $$g\!\left(\dfrac13\right)=\dfrac1{g(-1/3)}=\dfrac12.$$

Write $$-2=6\!\left(-\dfrac13\right)$$. Using the integer-multiple rule,

$$g(-2)=g\!\left(-\dfrac13\right)^{\,6}=2^{6}=64.$$

Now evaluate the required expression:

$$\bigl(f(1/4)+g(-2)-8\bigr)\,g(0)=\bigl((-5)+64-8\bigr)\,(1)=51.$

Hence the value is 51.