(a) $$f(x) = \sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right],\,x \in R$$

$$ = \sin \left( {{\pi \over 6}\sin \theta } \right),\,\theta \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$,

where, $$\theta = {\pi \over 2}\sin x$$

$$ = \sin \alpha ,\alpha \in \left[ { - {\pi \over 6},{\pi \over 6}} \right]$$,

where, $$\alpha = {\pi \over 6}\sin \theta $$

$$\therefore$$ $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$

Hence, range of $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$

So, option (a) is correct.

(b) $$f\{ g(x)\} = f(t),t \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$

$$ \Rightarrow f(t) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$

$$\therefore$$ Option (b) is correct.

(c) $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {g(x)}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 2}(\sin x)}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \right]} \over {{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)}}\,.\,{{{\pi \over 6}\sin \left( {{\pi \over 2}\sin x} \right)} \over {\left( {{\pi \over 2}\sin x} \right)}}$$

$$ = 1 \times {\pi \over 6} \times 1 = {\pi \over 6}$$

$$\therefore$$ Option (c) is correct.

(d) $$g\{ f(x)\} = 1$$

$$ \Rightarrow {\pi \over 2}\sin \{ f(x)\} = 1$$

$$ \Rightarrow \sin \{ f(x)\} = {2 \over \pi }$$ ..... (i)

But, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right] \subset \left[ { - {\pi \over 6},{\pi \over 6}} \right]$$

$$\therefore$$ $$\sin \{ f(x)\} \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$ ..... (ii)

$$ \Rightarrow \sin \{ f(x)\} \ne {2 \over \pi }$$, [from Eqs. (i) and (ii)]

i.e. No solution.

$$\therefore$$ Option (d) is not correct.