Answer
$h(x)$ is not in $span(f(x),g(x))$.
Work Step by Step
By definition, if $h(x) \in span(f(x),g(x))$ then there exists scalars $c_{1}$ and $c_{2}$ with
\[
h(x) = c_{1}f(x) + c_{2}g(x).
\]
That is,
\[
\sin x = c_{1} \sin^{2}x + c_{2} \cos^{2}x.
\]
As this must hold for every $x \in \mathbb{R}$, it in particular holds for $x_{1} =\frac{\pi}{2}$ and $x_{2}=0$. We thus have
\[
\sin \left(\frac{\pi}{2}\right)= c_{1} \sin^{2}\left(\frac{\pi}{2}\right) + c_{2} \cos^{2}\left(\frac{\pi}{2}\right)
\]
and
\[
\sin \left(0\right)= c_{1} \sin^{2}\left(0\right) + c_{2} \cos^{2}\left(0\right).
\]
That is, $1 = c_{1}\cdot 1 + c_{2} \cdot 0$ and $0 = c_{1} \cdot 0 + c_{2} \cdot 1$. Therefore, $c_{1} =1$ and $c_{2} =0$ giving
\[
\sin x = \sin^{2} x
\]
which is clearly false. Thus no such scalars $c_{1}$ and $c_{2}$ exist, hence $h(x) \notin span(f(x),g(x))$.
