Concept:
If a function is one–one and increasing, the equation \(f(x)=f^{-1}(x)\) represents intersection points of the graph of \(f(x)\) with the line \(y=x\).
Step 1: {Check monotonicity}
\[
f(x)=(x-1)^4+1
\]
\[
f'(x)=4(x-1)^3
\]
For \(x\ge1\),
\[
f'(x)\ge0
\]
Thus \(f(x)\) is increasing.
Step 2: {Solve \(f(x)=x\)}
\[
(x-1)^4+1=x
\]
\[
(x-1)^4=x-1
\]
Let \(y=x-1\).
\[
y^4=y
\]
\[
y(y^3-1)=0
\]
\[
y=0 \text{ or } y=1
\]
Thus
\[
x=1,\;2
\]
Hence Statement–1 is true.
Step 3: {Find inverse function}
\[
y=(x-1)^4+1
\]
\[
y-1=(x-1)^4
\]
\[
x-1=(y-1)^{1/4}
\]
\[
f^{-1}(x)=(x-1)^{1/4}+1
\]
Step 4: {Check Statement–2}
\[
f^{-1}(x+1)=(x+1-1)^{1/4}+1
\]
\[
=x^{1/4}+1
\]
Equation becomes
\[
x^{1/4}+1=(x-1)^4+1
\]
\[
x^{1/4}=(x-1)^4
\]
Graphically this equation has one solution, hence it is not true that it has no solution.
Thus Statement–2 is false.
Step 5: {Conclusion}
Statement–1 is true and Statement–2 is false.
\[
\boxed{\text{Option (3)}}
\]