Let $$f(x)=\dfrac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32}$$. Then the value of $$8\left(f\!\left(\dfrac{1}{15}\right)+f\!\left(\dfrac{2}{15}\right)+\cdots+f\!\left(\dfrac{59}{15}\right)\right)$$ is equal to:

$$f(x)=\dfrac{2^{x+2}+16}{2^{2x+1}+2^{x+4}+32} = \dfrac{2^{x+1}+8}{2^{2x}+8\cdot 2^{x}+16} = \dfrac{2(2^x+4)}{(2^x+4)^2}$$ 

Which gives,

$$f(x) = \dfrac{2}{2^x+4}$$

Putting $$4-x$$, we get,

$$f(4-x) = \dfrac{2}{2^{4-x}+4} = \dfrac{2\times 2^x}{2^4+4\times 2^x}$$

$$\Rightarrow f(4-x) = \dfrac{2^x}{2(2^x+4)}$$

Thus, $$f(x)+f(4-x) = \dfrac{1}{2}$$

We therefore get,

$$f\left(\dfrac{1}{15}\right) + f\left(\dfrac{59}{15}\right) = \dfrac{1}{2}$$

$$f\left(\dfrac{2}{15}\right) + f\left(\dfrac{58}{15}\right) = \dfrac{1}{2}$$

$$\vdots$$

$$f\left(\dfrac{29}{15}\right) + f\left(\dfrac{31}{15}\right) = \dfrac{1}{2}$$

With $$f\left(\dfrac{30}{15}\right)$$ remaining, which is simply equal to $$\dfrac{2}{2^{2}+4} = \dfrac{1}{4}$$

Therefore, the sum is equal to $$8\left(\dfrac{1}{4} + 29\times \dfrac{1}{2}\right) = 2 + 29\times 4 = 118$$