If f'(3) = 5 then $\lim\limits_{h \rightarrow 0} \frac{f\left(3+h^2\right)-f\left(3-h^2\right)}{2 h^2}$ is :

5

$\lim\limits_{h \rightarrow 0} \frac{f\left(3+h^2\right)-f\left(3-h^2\right)}{2 h^2}$

$=\lim\limits_{h \rightarrow 0} \frac{f\left(3+h^2\right)-f(3)+f(3)-f\left(3-h^2\right)}{2 h^2}$

$=\frac{1}{2} \lim\limits_{h \rightarrow 0} \frac{f\left(3+h^2\right)-f(3)}{h^2}+\frac{1}{2} \lim\limits_{h \rightarrow 0} \frac{f\left(3-h^2\right)-f(3)}{-h^2}$

$=\frac{1}{2} f'(3)+\frac{1}{2} f'(3)=\frac{1}{2}(5)+\frac{1}{2}(5)=5$

Hence (1) is correct answer.