The given curve is : $x=1+\frac{1}{y^2}$ ...(i)
$\Rightarrow 1=-2 \cdot \frac{1}{y^3} \cdot y^{\prime} \Rightarrow y^{\prime}=\frac{-y^2}{2}$
$\Rightarrow \quad\left(y^{\prime}\right)_{(2,1)}=-\frac{1}{2}$.
The equation of tangent at point $(2,1)$
$(y-1)=-\frac{1}{2}(x-2) \Rightarrow y-2=-\frac{1}{2} x$
$\Rightarrow \quad x=4-2 y$ ...(ii)
We can get the point $B$ by solving (i) and (ii)
$4-2 y=1+\frac{1}{y^2} \Rightarrow 2 y^3-3 y^2+1=0$
$\begin{aligned} & \Rightarrow \quad(y-1)^2(2 y+1)=0 \\ & \Rightarrow y=-\frac{1}{2}, 1,1\end{aligned}$
For point $B, y=-\frac{1}{2}$
At $y=-\frac{1}{2}, x=1+4=5$
$\therefore$ Co-ordinate of $B\left(5,-\frac{1}{2}\right)$
Let us find the equation of tangent at $B\left(5,-\frac{1}{2}\right)$


$\Rightarrow\left(y+\frac{1}{2}\right)=-\frac{1}{8}(x-5)$
$\Rightarrow y=-\frac{1}{8} x+\frac{1}{8}$ ...(iii)
$\because \quad$ Eqns. (ii) and (iii) are not same. So, option (1) is incorrect.
Now, let slope of tangent line at point $A$ and $B$ are $m_1$ and $m_2$ respectively. Then from equations (ii) and (iii), we get :
$m_1=-\frac{1}{2}, m_2=-\frac{1}{8}$
Angle between the tangents drawn at $A$ and $B$ is
$\theta=\tan ^{-1}\left(\frac{m_1+m_2}{1-m_1 m_2}\right)$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{-\frac{1}{2}-\frac{1}{8}}{1-\frac{1}{16}}\right)$
$\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{2}{3}\right) \neq 0$ or $\frac{\pi}{2}$