\(f(n)=2^n\), \(g(n)=n\), and \(h(n)=n^{\log n}\).
\[
\log f(n)
= \log\bigl(2^n\bigr)
= n\cdot\log 2
= \Theta(n).
\]
Also,
\[
\log h(n)
= \log\bigl(n^{\log n}\bigr)
= (\log n)^2.
\]
Therefore,
\[
g(n)
= n
= \Theta\bigl(\log f(n)\bigr).
\]
For \(h(n)\):
\[
\log_2 h(n)
= \log_2\bigl(n^{\log_2 n}\bigr)
= \log_2 n \cdot \log_2 n
= (\log_2 n)^2.
\]
For \(f(n) = 2^n\):
\[
\log_2 f(n)
= \log_2(2^n)
= n.
\]
Since for large \(n\), \((\log_2 n)^2 \ll n\),we get:
\[
\quad h(n) = O(2^n).
\]
\[
\boxed{\text{Option C is correct.}}
\]