Solutions:
Let \( h(x)=f(x)g'(x) \)
\( \Rightarrow \ h'(x)=f(x)g''(x)+f'(x)g'(x) \)
Hence, minimum no. of solutions of \( h'(x)=0 \) in \( (-2,2) \) are reqd.

\( f \) is even and \( f\!\left(\frac{1}{2}\right)=f\!\left(-\frac{1}{2}\right)=0 \)
\( \Rightarrow \ f\!\left(\frac{1}{4}\right)=f\!\left(-\frac{1}{4}\right)=0 \)
\( \Rightarrow \ h(x)=0 \) at \( x=\pm \frac{1}{4}, \pm \frac{1}{2} \)

\( g \) is even function \( \Rightarrow g(x)=g(-x)\ \forall x\in(-2,2) \)
\( \Rightarrow \ g'(x)=-g'(-x) \)
\( \Rightarrow \ g'(0)=g'(-0) \)
\( \Rightarrow \ g'(0)=0 \)
\( \Rightarrow \ h(0)=0 \)
\( \Rightarrow \ h \) has min. \( 5 \) roots in \( (-2,2) \)
\( \Rightarrow \ h'(x) \) has min. \( 4 \) roots in \( (-2,2) \)

And this minimum value is attained for \( f(x)=\frac{64}{45}\left(x^2-\frac{1}{4}\right)\left(x^2-\frac{1}{16}\right) \)

And \( g(x)=\frac{32}{7}\left(x^2-\frac{9}{16}\right) \)

\( h(x)=\frac{64}{7}\,x\left(x^2-\frac{1}{4}\right)\left(x^2-\frac{1}{16}\right) \)

Graphically or otherwise: \( h'(x) \) has exactly \( 4 \) roots in \( (-2,2) \) since it’s a polynomial of degree \( 4 \).
Graph of \( h(x) \):

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