"The formation of carbon monoxide is much favoured at higher temperature than formation of carbondioxide by direct union of carbon and oxygen" How can this be explained from thermodynamic point of view?

If you need some answer, I would suggest that perhaps because the bond length of the CO is about 3% shorter than CO2, and the bond energy is about 2% higher that CO2 bonds are more easily broken. This being the case and all else being equal, CO will be the favored product of C2+O2

According to your views CO is the favoured product of direct union of C and O as the bond energy of CO is higher than that of CO2.But the reason why it is favoured at higher temperature is not uderstood by your explanation.

C(s) + O2(g) = CO2 - 389kJ/mole
C(s) + 0.5O2(g) = CO - 110kJ/mole
In the first reaction there is no volume change
but in the second reaction the volume of the product is doubled that of reactants. So the change in entropy (dS) in first case is small but large in case of second reaction. Hence the free energy change (dG) [calculated according to equation-dG = dH - TdS]for the first reaction remains more or less constant with increase in temperaure in case of formation of CO2. But in case of second reaction the free energy change (dG) becmes more and more negative above 983K as T increases as evident from Eningham diagram. So the formation of CO will be favoured at higher temperature as the dG becoming more and more negative with increase of temperature.