Suppose the $64$ distinct positive real numbers are $x_1<x_2<...<x_{64}$ Let $a_1>a_2$ denote two largest numbers written on red cards. Let $b_1>b_2$ denote two largest numbers written on blue cards. Note that $a_1\neq a_2$ and $b_1\neq b_2$ because $x_{63}+x_{64}\neq x_{62}+x_{64}$ and $x_{64}x_{63}\neq x_{64}x_{62}$ Then compare the values $\frac{a_1}{a_2}$ and $\frac{b_1}{b_2}$. Since $\frac{x_{64}+x_{63}}{x_{64}+x_{62}} < \frac{x_{64}x_{63}}{x_{64}x_{62}}$. So the larger value must come from the cards correspond to product of $x_i$s

Or alternatively, suppose that such two largest numbers can be both sum and product of the three largest real numbers, i.e. $a_1 = x_{63} + x_{64} = y_{63} y_{64}$, $a_2 = x_{62} + x_{64} = y_{62} y_{64}$ $b_1 = y_{63} + y_{64} = x_{63} x_{64}$, $b_2 = y_{62} +y_{64} = x_{62} x_{64}$ Solve the equations for $x_{64}$ and $y_{64}$, like $${x_{64}}^2 - a_1 x_{64} + b_1 = 0, {x_{64}}^2 - a_2 x_{64} + b_2 =0$$$${y_{64}}^2 - b_1 y_{64} + a_1 = 0, {y_{64}}^2 - b_2 y_{64} + a_2 =0$$and you can find that $a_1 = a_2$, contradiction.