$a)f\left(x\right)=\frac{x^2-4x+1}{x^2}=1-\frac{4}{x}+\frac{1}{x^2}$  

$f^{{}^{\prime }}\left(x\right)=\left(1\right){}^{\prime }-{\left(\frac{4}{x}\right)}^{{}^{\prime }}+{\left(\frac{1}{x^2}\right)}^{{}^{\prime }}=-\left(-\frac{4}{x^2}\right)+\frac{\left(1\right){}^{\prime }\left(x^2\right)-\left(1\right)\left(x^2\right){}^{\prime }}{x^4}=\frac{4}{x^2}+\frac{-2x}{x^4}=\frac{4x^2}{x^4}-\frac{2x}{x^4}=\frac{4x^2-2x}{x^4}$ $f^{{}^{\prime }}\left(x\right)=\left(1\right){}^{\prime }-{\left(\frac{4}{x}\right)}^{{}^{\prime }}+{\left(\frac{1}{x^2}\right)}^{{}^{\prime }}=-\left(-\frac{4}{x^2}\right)+\frac{\left(1\right){}^{\prime }\left(x^2\right)-\left(1\right)\left(x^2\right){}^{\prime }}{x^4}=\frac{4}{x^2}+\frac{-2x}{x^4}=\frac{4x^2}{x^4}-\frac{2x}{x^4}=\frac{4x^2-2x}{x^4}$  

 

$f^{{}^{\prime }}\left(1\right)=\mathrm{tg}\alpha =a$  

$f^{{}^{\prime }}\left(1\right)=\frac{4-2}{1}=2$  

$a=2$  

 

 

$b)f\left(x\right)=\frac{3x^2-1}{1-x^2}=\frac{x^2-1+2x^2}{1-x^2}=\frac{x^2-1}{1-x^2}+\frac{2x^2}{1-x^2}=\frac{2x^2}{1-x^2}-1$  

$f^{{}^{\prime }}\left(x\right)={\left(\frac{2x^2}{1-x^2}\right)}^{{}^{\prime }}-{\left(1\right)}^{{}^{\prime }}=\frac{{\left(2x^2\right)}^{{}^{\prime }}\left(1-x^2\right)-\left(2x^2\right){\left(1-x^2\right)}^{{}^{\prime }}}{{\left(1-x^2\right)}^2}=\frac{4x\left(1-x^2\right)-\left(-2x\right)\left(2x^2\right)}{{\left(1-x^2\right)}^2}=\frac{4x-4x^3+4x^3}{{\left(1-x^2\right)}^2}=\frac{4x}{{\left(1-x^2\right)}^2}$  

 

$f^{{}^{\prime }}\left(2\right)=\mathrm{tg}\alpha =a$  

$f^{{}^{\prime }}\left(2\right)=\frac{8}{9}$