Experts kindly help me solve these two questions.

Dear Student,
1. Given,
$$\begin{gathered} Angle of incidence, i_1={48}^o \\ Refracting angle or angle of prism, A = {60}^o \\ Let the angle of deviation be \delta , then \\ \delta =\left(i_1+i_2\right)-A \\ where, i_2 is the angle of emergence \\ At minimum deviation \left({\delta }_m\right), angle of incidence = angle of emergence \\ {\delta }_m=2i_1-A=2\times {48}^o-{60}^o={36}^o. \end{gathered}$$
2.  Given,
Angle of minimum deviation to be 36^o. 
Prism is an equilateral prism, i.e., A = 60^o.
$$\begin{gathered} So, \\ {\delta }_m={36}^o; A={60}^o \\ U\sin g the relation \\ {\delta }_m=2i_1-A \\ where, i_1 is the angle of incidence \\ {i}_1=\frac{\left(A+{\delta }_m\right)}{2}=\frac{{60}^o+{36}^o}{2}={48}^o \end{gathered}$$

Regards