If two distinct tangents can be drawn from the point $(\mathrm{\alpha },2)$ on different branches of the hyperbola  $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{16}=1$, then 

$$\begin{gathered} For hyperbola \frac{x^2 }{9}-\frac{y^2}{16}= 1 \\ to have two distinct \tan gents on different branches \\ the point should lie in the region R_1 and R_3 \\ So from ( \alpha , 2 ) two \tan gents can be drawn \\ if it lies in between A and B [where A and B are point of intersection asymptotes with y = 2 \\ Equation of pair of assymptotes is\frac{x^2 }{9}-\frac{y^2}{16}=0 \\ \Rightarrow (\frac{x}{3}-\frac{y}{4}\left)\right(\frac{x}{3}+\frac{y}{4}) \\ \Rightarrow (4x-3y\left)\right(4x+3y)=0 \end{gathered}$$

Equation of asymptotes are $4\mathrm{x}=\pm 3\mathrm{y}$.

Solving with $\mathrm{y}=2$, we have $\mathrm{x}=\pm \frac{3}{2}$

$$\begin{gathered} -\frac{3}{2}<\mathrm{\alpha }<\frac{3}{2} \\ \Rightarrow \left|\alpha \right|<\frac{3}{2} \end{gathered}$$