a)
[tex]\bf y=x\left( 1-\cfrac{2}{x}+x \right)\implies y=x-2+x^2
\\\\\\
\cfrac{dy}{dx}=1-0+2x\implies \left. \cfrac{dy}{dx}=1+2x \right|_{x=1}\implies \stackrel{m}{3}
\\\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad
\begin{cases}
x=1\\
y=0\\
m=3
\end{cases}\implies y-0=3(x-1)
\\\\\\
y=3x-1[/tex]
b)
recall that the is the equation of the tangent line at (1,0), whose slope is 3 of course, now, the normal will be the line that's perpendicular to the tangent line, and therefore it will have a slope that is negative reciprocal to the tangen't line's slope, thus
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad 3\implies \cfrac{3}{1}\\\\
negative\implies -\cfrac{3}{ 1}\qquad reciprocal\implies - \cfrac{ 1}{3}\\\\
-------------------------------\\\\
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\quad
\begin{cases}
x=1\\
y=0\\
m=-\frac{1}{3}
\end{cases}\implies y-0=-\cfrac{1}{3}(x-1)
\\\\\\
y=-\cfrac{1}{3}x+\cfrac{1}{3}[/tex]
now, for the second bit there,
[tex]\bf sin\left( \sqrt{x^2+1} \right)\implies sin\left[ ( x^2+1 )^{\frac{1}{2}} \right]
\\\\\\
\stackrel{chain~rule}{cos\left[ ( x^2+1 )^{\frac{1}{2}} \right]\cdot \cfrac{1}{2}( x^2+1 )^{-\frac{1}{2}}\cdot 2x}
\\\\\\
cos\left[ ( x^2+1 )^{\frac{1}{2}} \right]\cdot \cfrac{x}{( x^2+1 )^{\frac{1}{2}}}\implies \cfrac{x~cos\left( \sqrt{x^2+1}\right)}{\sqrt{x^2+1}}[/tex]
