Given function

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

First simplify the expression inside the power.

Simplify the inside

$2\left(1-\frac{x^{25}}{2}\right)=2-x^{25}$

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

Using difference of squares, $(2-x^{25})(2+x^{25})=4-x^{50}$

Hence, $f(x)=\left(4-x^{50}\right)^{1/50}$

Compute $f(1)$

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

Compute $f(f(1))$

Let $a=f(1)=3^{1/50}$. Then $a^{50}=3$

Therefore, $f(f(1))=f(a)=\left(4-a^{50}\right)^{1/50}=\left(4-3\right)^{1/50}=1$

Compute $f(f(f(1)))$

Since $f(f(1))=1$, $f(f(f(1)))=f(1)=3^{1/50}$

Now evaluate $g(1)$

$g(x)=f(f(f(x)))+f(f(x))$

So, $g(1)=f(f(f(1)))+f(f(1))=3^{1/50}+1$

Find the greatest integer less than or equal to $g(1)$

Since $3^{1/50}>1$ and clearly $3^{1/50}<2,$ we get $2<1+3^{1/50}<3$

Therefore, $\lfloor g(1)\rfloor=2$

Comparison with stored answer

Derived answer = $2$

Stored correct answer = $2$

They match.