Suppose f is a function satisfying f ( x + y ) = f ( x ) + f ( y )...

Given,

$f\left(x + y\right) = f\left(x\right) + f\left(y\right) \text{ and } f\left(1\right) = \dfrac{1}{5}$

Now solving the given functional equation by taking $x = 1 \text{ and } y = 1$ we get,

$f\left(1 + 1\right) = f\left(1\right) + f\left(1\right)$

$⇒ f\left(2\right) = \dfrac{2}{5}\left\{\text{as}f\left(1\right) = \dfrac{1}{5}\right\}$

Now taking $x = 2 \text{ and } y = 1$ we get,

$f\left(3\right) = \dfrac{2}{5} + \dfrac{1}{5} ⇒ f\left(3\right) = \dfrac{3}{5}$

Hence, $f\left(x\right) = \dfrac{x}{5}$

Now solving $\mathop{∑}_{n = 1}^{m}\dfrac{f\left(n\right)}{n\left(n + 1\right)\left(n + 2\right)} = \dfrac{1}{12}$

$⇒ \mathop{∑}_{n = 1}^{m}\dfrac{\dfrac{n}{5}}{n\left(n + 1\right)\left(n + 2\right)} = \dfrac{1}{12}$

$⇒ \dfrac{1}{5}\mathop{∑}_{n = 1}^{m}\dfrac{1}{\left(n + 1\right)\left(n + 2\right)} = \dfrac{1}{12}$

$⇒ \dfrac{1}{5}\mathop{∑}_{n = 1}^{m}\dfrac{1}{\left(n + 1\right)} - \dfrac{1}{\left(n + 2\right)} = \dfrac{1}{12}$

$⇒ \dfrac{1}{5}\left[\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} . . . . . - \dfrac{1}{m + 2}\right] = \dfrac{1}{12}$

$⇒ \dfrac{1}{5}\left[\dfrac{1}{2} - \dfrac{1}{m + 2}\right] = \dfrac{1}{12}$

$⇒ \left[\dfrac{m}{2\left(m + 2\right)}\right] = \dfrac{5}{12}$

$⇒ 12m = 10\left(m + 2\right)$

$⇒ m = 10$