Since at equilibrium, Moles of CO= Moles of Zn= moles of CO2 (as no information about initial moles are given)
If no information is given, you should just give the three amounts names and treat them as unknowns.
ZnO is exposed to pure CO at 1300 K
This means that initially there is no carbon dioxide and no elemental zinc. Carbon dioxide and elemental zinc will be present in a 1:1 ratio because that is the ratio they are produced at.
The density of the gas mixture is 0.344 g/L at the same temperature.
Let's find the density of pure CO at 1300 K and 1 atm pressure, assuming it behaves as ideal gas.
$$\rho_1 = m/V = M n / V = M P / (R T) = $$ $$\frac{\pu{28.01 g/mol} \cdot \pu{1 atm}}{\pu{0.08206 L atm K-1 mol-1}\cdot \pu{1300 K}} = \pu{0.2626 g/L}$$
And now the density of a 1:1 mixture of carbon dioxide and zinc under the same conditions. The average molar mass of the two species is 0.5 (65.4 + 44.01) g/mol.
$$\rho_2 = m/V = M_\mathrm{ave} n / V = M P / (R T) = $$ $$\frac{\pu{54.71 g/mol} \cdot \pu{1 atm}}{\pu{8.314 J K-1 mol-1}\cdot \pu{1300 K}} = \pu{0.5129 gm3 g/L}$$
So the density of the gas for pure reactants is lower than the observed density, and the density for pure products is higher. The equilibrium mixture contains some reactant and some product, and the density is a weighted average of the two densities just calculated (with $X_\ce{CO}$ the mole fraction of carbon monoxide in the gas phase of the reaction mixture):
$$\rho = X_\ce{CO} \rho_1 + (1-X_\ce{CO}) \rho_2$$
We can solve for $X_\ce{CO}$ and calculate its value:
$$\rho - \rho_2 = X_\ce{CO} (\rho_1 - \rho_2)$$
$$ X_\ce{CO} = \frac{\rho - \rho_2}{\rho_1 - \rho_2} = 0.675 $$
So the partial pressure of carbon monoxide is 0.675 atm.