$$\begin{aligned} & \mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \quad \text{... (1)}\\ & \Rightarrow \quad \mathrm{f}(\mathrm{nx})=\operatorname{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{N} \quad \text{... (2)} \end{aligned}$$

$$ \begin{array}{ll} \text { Now } & \text { put } \mathrm{y}=-\mathrm{x} \text { in eq.(1) } \\ & \mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})=\mathrm{f}(0) \quad\{\mathrm{f}(0)=0\} \\ \Rightarrow \quad & \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \\ \Rightarrow \quad & \mathrm{f} \text { is odd function } \\ & \text { from eq. (2) } \\ & \mathrm{f}(-\mathrm{nx})=\mathrm{nf}(-\mathrm{x}) \\ \Rightarrow \quad & \mathrm{f}(-\mathrm{nx})=-\mathrm{nf}(\mathrm{x}) \\ \Rightarrow \quad & \mathrm{f}(\mathrm{mx})=\operatorname{mf}(\mathrm{x}) \forall \mathrm{m} \in \mathrm{Z}^{-} \quad \text{... (3)}\\ & \text {from eq. (2) and eq. (3) } \\ & \mathrm{f}(\mathrm{nx})=\mathrm{nf}(\mathrm{x}) \forall \mathrm{n} \in \mathrm{Z} \quad \text{... (4)} \end{array}$$

$$\begin{aligned} \text { Now } & \text { put } x=\frac{p}{q} \text { where } p, q \in Z, q \neq 0 \\ f\left(\frac{n p}{q}\right) & =n f\left(\frac{p}{q}\right) \forall n \in Z \\ \text { put } n & =q \\ & f(p)=q f\left(\frac{p}{q}\right) \end{aligned}$$

$$\Rightarrow \quad \mathrm{pf}(1)=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \quad \{\text{from eq.(4)}\}$$

Let $$f(1)=a$$

$$\begin{array}{ll} \text { then } \quad & \mathrm{pa}=\mathrm{qf}\left(\frac{\mathrm{p}}{\mathrm{q}}\right) \\ & \mathrm{f}\left(\frac{\mathrm{p}}{\mathrm{q}}\right)=\frac{\mathrm{ap}}{\mathrm{q}} \\ \Rightarrow \quad & \mathrm{f}(\mathrm{x})=\mathrm{ax} \forall \mathrm{x} \in \mathbb{Q} \end{array}$$

Now, $$f\left(\frac{-3}{5}\right)=a\left(\frac{-3}{5}\right)=12 \Rightarrow a=-20$$

$$\Rightarrow \quad \mathrm{f}(\mathrm{x})=-20 \mathrm{x} \forall \mathrm{x} \in \mathbb{Q} \quad \text{... (5)}$$

From the given functional equation it is not possible to find a unique function for irrational values of '$$x$$', there are infinitely many such functions satisfying given functional equation for irrational values of $$x$$, but in this problem we finally need the function at rational values of '$$x$$' only. So, for rational values of $$x$$ we are getting a unique function mentioned in (5).

Now, $$g(x+y)=g(x) \cdot g(y)$$

$$\begin{array}{ll} \Rightarrow & \ln (\mathrm{g}(\mathrm{x}+\mathrm{y})=\ln (\mathrm{g}(\mathrm{x}))+\ln (\mathrm{g}(\mathrm{y})) \\ \text { Let } & \ln (\mathrm{g}(\mathrm{x}))=\mathrm{h}(\mathrm{x}) \\ \Rightarrow & \mathrm{h}(\mathrm{x}+\mathrm{y})=\mathrm{h}(\mathrm{x})+\mathrm{h}(\mathrm{y}) \\ \Rightarrow & \mathrm{h}(\mathrm{x})=\mathrm{kx} \forall \mathrm{x} \in \mathbb{Q} \\ \Rightarrow & \mathrm{g}(\mathrm{x})=\mathrm{e}^{\mathrm{kx}} \forall \mathrm{x} \in \mathbb{Q} \quad \text{... (6)} \end{array}$$

$$\begin{aligned} & \text { and } \quad \mathrm{g}\left(\frac{-1}{3}\right)=\mathrm{e}^{-\frac{\mathrm{K}}{3}}=2 \quad \Rightarrow \quad \mathrm{K}=-3 \ln 2 \\ & \Rightarrow \quad \mathrm{K}=\ln \left(\frac{1}{8}\right) \\ & \Rightarrow \quad \mathrm{g}(\mathrm{x})=\mathrm{e}^{\ln \left(\frac{1}{8}\right) \cdot \mathrm{x}}=\left(\frac{1}{8}\right)^x=2^{-3 \mathrm{x}} \forall \mathrm{x} \in \mathbb{Q} \end{aligned}$$

Now, $$f\left(\frac{1}{4}\right)=-5, g(-2)=2^6=64$$

$$\mathrm{g}(0)=1$$

$$\begin{aligned} \text{So} \quad & \left(\mathrm{f}\left(\frac{1}{4}\right)+\mathrm{g}(-2)-(8) \mathrm{g}(0)\right) \\ & =(-5+64-8)(1)=51 \end{aligned}$$