Given,

$$f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)\left( {2 + {x^{25}}} \right)} \right)^{{1 \over {50}}}}$$

and $$g(x) = f\left( {f\left( {f\left( x \right)} \right)} \right) + f\left( {f\left( x \right)} \right)$$

$$\therefore$$ $$g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$$

Now, $$f(1) = {\left( {2\left( {1 - {{{1^{25}}} \over 2}} \right)\left( {2 + {1^{25}}} \right)} \right)^{{1 \over {50}}}}$$

$$ = {\left( {2\left( {1 - {1 \over 2}} \right)\left( {2 + 1} \right)} \right)^{{1 \over {50}}}}$$

$$ = {\left( 3 \right)^{{1 \over {50}}}}$$

$$\therefore$$ $$f\left( {f\left( 1 \right)} \right) = f\left( {{3^{{1 \over {50}}}}} \right)$$

$$ = {\left( {2\left( {1 - {{{{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \over 2}} \right)\left( {2 + {{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \right)} \right)^{{1 \over {50}}}}$$

$$ = {\left( {2\left( {1 - {{{3^{{1 \over 2}}}} \over 2}} \right)\left( {2 + {3^{{1 \over 2}}}} \right)} \right)^{{1 \over {50}}}}$$

$$ = {\left( {2 \times \left( {{{2 - \sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)} \right)^{{1 \over {50}}}}$$

$$ = {\left[ {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} \right]^{{1 \over {50}}}}$$

$$ = {\left( {4 - 3} \right)^{{1 \over {50}}}}$$

$$ = {1^{{1 \over {50}}}} = 1$$

Now, $$f\left( {f\left( {f\left( 1 \right)} \right)} \right) = f(1) = {3^{{1 \over {50}}}}$$

$$\therefore$$ $$g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$$

$$ = {3^{{1 \over {50}}}} + 1$$

Now, greatest integer less than or equal to $$g(1)$$

$$ = \left[ {g(1)} \right]$$

$$ = \left[ {{3^{{1 \over {50}}}} + 1} \right]$$

$$ = \left[ {{3^{{1 \over {50}}}}} \right] + \left[ 1 \right]$$

$$ = [1.02] + 1$$

$$ = 1 + 1 = 2$$