For the function $f:[1,\infty) \rightarrow [1,\infty)$ defined by $f(x)=(x-1)^4+1$, among the two statements:
(I) The set $S=\{x \in [1,\infty): f(x)=f^{-1}(x)\}$ contains exactly two elements, and
(II) The set $S=\{x \in [1,\infty): f(x)=f^{-1}(x+1)\}$ is an empty set,

Given $f(x) = (x-1)^4 + 1$ for $x \ge 1$.

To find the inverse function $f^{-1}(x)$, let $y = (x-1)^4 + 1$.

$\Rightarrow (x-1)^4 = y-1 \Rightarrow x-1 = (y-1)^{1/4} \Rightarrow x = (y-1)^{1/4} + 1$.

Thus, $f^{-1}(x) = (x-1)^{1/4} + 1$.

Evaluating the first statement, since $f'(x) = 4(x-1)^3 \ge 0$ for $x \ge 1$, $f(x)$ is strictly increasing. For a strictly increasing function, the solutions to $f(x) = f^{-1}(x)$ are the same as the solutions to $f(x) = x$.

$(x-1)^4 + 1 = x \Rightarrow (x-1)^4 - (x-1) = 0$.

Let $t = x-1 \ge 0$.

$t^4 - t = 0 \Rightarrow t(t-1)(t^2+t+1) = 0$.

Since $t \ge 0$, the real roots are $t = 0$ and $t = 1$.

$\Rightarrow x-1 = 0 \Rightarrow x = 1$

$\Rightarrow x-1 = 1 \Rightarrow x = 2$

Both $x=1$ and $x=2$ lie in $[1, \infty)$. The set contains exactly two elements, making Statement (I) TRUE.

Evaluating the second statement, the equation is $f(x) = f^{-1}(x+1)$.

$f^{-1}(x+1) = (x+1-1)^{1/4} + 1 = x^{1/4} + 1$.

Equating $f(x)$ and $f^{-1}(x+1)$:

$(x-1)^4 + 1 = x^{1/4} + 1 \Rightarrow (x-1)^4 - x^{1/4} = 0$.

Let $g(x) = (x-1)^4 - x^{1/4}$.

$g(2) = (2-1)^4 - 2^{1/4} = 1 - 2^{1/4} < 0$.

$g(3) = (3-1)^4 - 3^{1/4} = 16 - 3^{1/4} > 0$.

Since $g(x)$ is a continuous function on $[1, \infty)$ and changes sign between $x=2$ and $x=3$, by the Intermediate Value Theorem, there exists at least one real root in the interval $(2, 3)$.

Thus, the set is not empty, making Statement (II) FALSE.

Therefore, only (I) is TRUE.

Answer: only (I) is TRUE