$$\begin{gathered} \text{If}x\in [\frac{-\sqrt{3}}{2},\frac{1}{2}]\text{, then maximum value of the expression}\left({\sin ^-1x}^2\right)+\left({\cos ^-1x}^2\right)\text{is }\frac{n{\pi }^2}{n+7}\text{ (where}n\in N)\text{, then}\mathrm{ n}\text{ is equal to} \\ \end{gathered}$$

The correct answer is 29.

We are given that $x\in [\frac{-\sqrt{3}}{2},\frac{1}{2}]$.
Let us analyze the expression $E\left(x\right)=({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2$.

We know the standard identity for inverse trigonometric functions:
${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

Using the algebraic identity $a^2+b^2=(a+b{)}^2-2ab$, where $a={\sin }^{-1}x$ and $b={\cos }^{-1}x$:
$E\left(x\right)=({\sin }^{-1}x+{\cos }^{-1}x{)}^2-2{\sin }^{-1}x·{\cos }^{-1}x$
$E\left(x\right)={\left(\frac{\pi }{2}\right)}^2-2{\sin }^{-1}x\left(\frac{\pi }{2}-{\sin }^{-1}x\right)$
$E\left(x\right)=\frac{{\pi }^2}{4}-\pi {\sin }^{-1}x+2({\sin }^{-1}x{)}^2$

Let $t={\sin }^{-1}x$.
Since $x\in [\frac{-\sqrt{3}}{2},\frac{1}{2}]$, the range of $t$ is:
$t\in \left[{\sin }^{-1}\left(\frac{-\sqrt{3}}{2}\right),{\sin }^{-1}\left(\frac{1}{2}\right)\right]=\left[-\frac{\pi }{3},\frac{\pi }{6}\right]$

Now we write $E\left(t\right)$ as a quadratic function in terms of $t$:
$E\left(t\right)=2t^2-\pi t+\frac{{\pi }^2}{4}$
Completing the square for $E\left(t\right)$:
$E\left(t\right)=2\left(t^2-\frac{\pi }{2}t\right)+\frac{{\pi }^2}{4}$
$E\left(t\right)=2{\left(t-\frac{\pi }{4}\right)}^2-2{\left(\frac{\pi }{4}\right)}^2+\frac{{\pi }^2}{4}$
$E\left(t\right)=2{\left(t-\frac{\pi }{4}\right)}^2+\frac{{\pi }^2}{8}$

This is a parabola opening upwards with its vertex at $t=\frac{\pi }{4}$. Since the vertex $t=\frac{\pi }{4}$ lies outside the given interval $\left[-\frac{\pi }{3},\frac{\pi }{6}\right]$, the maximum value of the function on the interval occurs at the endpoint furthest from $t=\frac{\pi }{4}$.

Comparing the values of $E\left(t\right)$ at the boundaries:
At $t=\frac{\pi }{6}$:
$E\left(\frac{\pi }{6}\right)=2{\left(\frac{\pi }{6}\right)}^2-\pi \left(\frac{\pi }{6}\right)+\frac{{\pi }^2}{4}=\frac{{\pi }^2}{18}-\frac{{\pi }^2}{6}+\frac{{\pi }^2}{4}=\frac{2{\pi }^2-6{\pi }^2+9{\pi }^2}{36}=\frac{5{\pi }^2}{36}$

At $t=-\frac{\pi }{3}$:
$E\left(-\frac{\pi }{3}\right)=2{\left(-\frac{\pi }{3}\right)}^2-\pi \left(-\frac{\pi }{3}\right)+\frac{{\pi }^2}{4}=\frac{2{\pi }^2}{9}+\frac{{\pi }^2}{3}+\frac{{\pi }^2}{4}=\frac{8{\pi }^2+12{\pi }^2+9{\pi }^2}{36}=\frac{29{\pi }^2}{36}$

Thus, the maximum value of the expression is $\frac{29{\pi }^2}{36}$.

We are given that the maximum value is of the form:
$\frac{n{\pi }^2}{n+7}$

Equating the two expressions for the maximum value:
$\frac{n{\pi }^2}{n+7}=\frac{29{\pi }^2}{36}$
$\frac{n}{n+7}=\frac{29}{36}$
$36n=29(n+7)$
$36n=29n+203$
$7n=203$
$n=29$

Hence, $n$ is equal to 29.