Given, $$f(x + y) = f(x)\,.\,f(y)\,\forall x,y \in R$$

$$\therefore$$ $$f(x) = {a^x} \Rightarrow f'(x) = {a^x}\,.\,\log (a)$$

Now, $$f'(0) = \log (a) \Rightarrow 3 = \log (a) \Rightarrow a = {e^3}$$

$$\therefore$$ $$f(x) = {({e^3})^x} = {e^{3x}}$$

$$\therefore$$ $$f(h) = {e^{3h}}$$

Now, $$\mathop {\lim }\limits_{h \to 0} \left( {{{f(h) - 1} \over h}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over h}} \right)$$

$$ = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over {3h}} \times 3} \right) = 3 \times 1 = 3$$