You are given the sample mean and

Solving a simple quadratic in modular arithmetic
$x^2\equiv 9\pm \mathrm{mod}13$

Given ${\displaystyle \cot \left(b\right)=-2}$ find ${\displaystyle \sin \left(4b\right)}$ and ${\displaystyle \cos \left(4b\right)}$
I made $\sin \left(4b\right)$ into a expanded form.
${\displaystyle 4\sin \left(b\right){\cos }^3\left(b\right)-4{\sin }^3\left(b\right)\cos \left(b\right)}$ And then I made a triangle using ${\displaystyle \cot \left(b\right)=-2}$for information.
I got from that triangle that, ${\displaystyle \sin \left(b\right)=\frac{\sqrt{5}}{5}}$ and that ${\displaystyle \cos \left(b\right)=\frac{-2\sqrt{5}}{5}}$
And from their I simplified.
But whenever I expand ${\displaystyle \cos 4b}$ and plug in I get a bad answer, ${\displaystyle \frac{353}{25}}$ and I do not know if it is correct or wrong.

How to solve $\mathbf{\text{x}}\text{'}\text{'}\left(t\right)=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\mathbf{\text{x}}\left(t\right)$
What I'm thinking is to consider the first order version
$\mathbf{\text{x}}\text{'}\left(t\right)=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\mathbf{\text{x}}\left(t\right)$
which I know how to solve, the solution is
$\mathbf{\text{x}}\left(t\right)=\mathbf{\text{c}}\left[\begin{array}{c}{e}^{-t}\\ e^{-t}\end{array}\right]$
How do I use this to solve the second-order equation?

What is the equation of the line that is normal to $f\left(x\right)=-3x^2-\sin x$ at ${\displaystyle x=\frac{\pi }{3}}$?

Is there a way to evaluate this limit:
$\underset{x\to 0}{lim}\frac{\sin (e^{{\tan }^{2}x}-1)}{{\cos }^{\frac{3}{5}}\left(x\right)-\cos \left(x\right)}$

I have
$\frac{\tan 15°}{1-\tan {{15}^{\circ }}^2}$