From the table of normal shock properties; for, Mach number of $2.8$, $\frac{{{p_2}}}{{{p_1}}} = 8.98$, $\frac{{{\rho _2}}}{{{\rho _1}}} = 3.664$, $\frac{{{T_2}}}{{{T_1}}} = 2.451$, $\frac{{{p_{02}}}}{{{p_1}}} = 10.57$, ${M_2} = 0.4882$.

Therefore,

$${p_2} = 8.98{p_1} = 8.98\left( {1atm} \right) = 8.98\,atm$$

$${T_2} = 2.451{T_1} = 2.451\left( {295} \right) = 723.045\,K$$

$${p_{_1}} = {\rho _1}R{T_1} \Rightarrow {\rho _1} = \left( {\frac{{1 \times \left( {1.01 \times {{10}^5}} \right)}}{{287 \times 295}}} \right) = 1.193\,kg/{m^3}$$

$${\rho _2} = 3.664\left( {1.193} \right) = 4.371\,kg/{m^3}$$

$${p_{02}} = \left( {\frac{{{p_{02}}}}{{{p_1}}}} \right){p_1} = 10.57\left( 1 \right) = 10.57\,atm$$

From isentropic flow properties table, for, Mach number of  $2.8$,  $$\frac{{{T_{01}}}}{{{T_1}}} = 2.568$$

Since, ${T_{02}} = {T_{01}}$, Total temperature is constant,

Therefore,

$${T_{02}} = \left( {\frac{{{T_{01}}}}{{{T_1}}}} \right){T_1} = 2.568 \times \left( {295} \right) = 757.56\,K$$