If 8 and 2 are the roots of $x^2+ax+\beta =0$ and 3, 3 are the roots of $x^2+\alpha x+b=0$,  then the roots of the equation $x^2+ax+b=0$ are 

Given $8,2$ are the roots of the equation $x^2+ax+\beta =0$

Hence, $a=-\left(8+2\right)=-10$

Given $3,3$ are the roots of the equation $x^2+\alpha x+b=0$

Hence, $b=9 ( product of the roots)$

The required equation is $x^2-10x+9=0$

Factorize the above equation as 

$\begin{array}{rcl}{x}^2-x-9x+9& =& 0\\ x\left(x-1\right)-9\left(x-1\right)& =& 0\\ \left(x-1\right)\left(x-9\right)& =& 0\end{array}$

It implies that 

$\begin{array}{rcl}x-1& =& 0\\ x& =& 1 or\\ x-9& =& 0\\ x& =& 9\end{array}$

Therefore, required roots are $\boxed{1,9}$,