Given, $$f\left( {g(x)} \right) = x$$

$$g(x) = {f^{ - 1}}(x)$$

$$ \Rightarrow {d \over {dx}}g(x) = {d \over {dx}}{f^{ - 1}}(x)$$

$$ \Rightarrow {d \over {dx}}{f^{ - 1}}{(x)_{at\,x = 63}} = {1 \over {{d \over {dx}}f{{(x)}_{at\,x = 4}}}}$$

$$ = {1 \over {{{(3{x^2} + 1)}_{at\,x = 4}}}}$$

$$ = {1 \over {49}}$$

A = 4-digit numbers divisible by 3

A = 1002, 1005, ....., 9999.

9999 = 1002 + (n $$-$$ 1)3

$$\Rightarrow$$ (n $$-$$ 1)3 = 8997 $$\Rightarrow$$ n = 3000

B = 4-digit numbers divisible by 7

B = 1001, 1008, ......., 9996

$$\Rightarrow$$ 9996 = 1001 + (n $$-$$ 1)7

$$\Rightarrow$$ n = 1286

A $$\cap$$ B = 1008, 1029, ....., 9996

9996 = 1008 + (n $$-$$ 1)21

$$\Rightarrow$$ n = 429

So, no divisible by either 3 or 7

= 3000 + 1286 $$-$$ 429 = 3857

total 4-digits numbers = 9000

required numbers = 9000 $$-$$ 3857 = 5143

A = ($$-$$$$\infty$$, 1) $$\cup$$ (3, $$\infty$$)

B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)

C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)

So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)

z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}

Hence, no. of its subsets = 2⁸ = 256.