Taken from Khan Academy:
The nth derivative of $g$ at $x=0$ is given by $g^{(n)}(0)=\cfrac{\sqrt{n+7}}{n^3}$ for $n\geq1$
What is the coefficient for the term containing $x^2$ in the Maclaurin series of $g$?
The answer says: The quadratic term of the Taylor series centered at $x=0$ is $g^{\prime\prime}(0)\cfrac{x^2}{2!}$
From the given information, $g^{\prime\prime}(0)=\cfrac{\sqrt{9}}{8}=\cfrac{3}{8}$
Then the coefficient of $x^2$ is $\cfrac{3}{8}\cdot\cfrac{1}{2!}=\cfrac{3}{16}$
Why do they use two substituted for the value of n?
The notation $g^{(n)}(x)$ refers to the $n$-th derivative of $g(x)$. Thus, $g''(0)=g^{(2)}(0)$, and (borrowing the nice formatting from Holo in a comment)
$$ g^{(n)}(0)=g{\overbrace{'\!\!\cdots'}^{\text{n times}}}(0)\;. $$