If $x=2$and $x=3$ are the roots of the equation $3x^2-2kx+2m=0$ the values of  k and m are 

2, 3 are roots of $3x^2-2kx+2m=0$

Sum:$2+3=\frac{-(-2k)}{3}$

$5=\frac{2k}{3}\Rightarrow k=\frac{15}{2}$

Product:$2\times 3=\frac{2m}{3}$

$6=\frac{2m}{3}\Rightarrow m=3\times 3=9$

$(k,m)=(\frac{15}{2},9)$