HOW TO SOLV THIS PROBLEM

According to the law of conservation of mass, the total mass of reactants equals the total mass of products.
Mass of metal (M) + Mass of oxygen ($\ce{O2}$) = Mass of metal oxide (MO)
Mass of $\ce{O2}$ = Mass of MO - Mass of M
Mass of $\ce{O2}$ = 1.27 g - 1.00 g
Mass of $\ce{O2}$ = 0.27 g

The mass of oxygen that reacted is 0.27\ g.

Molar mass of O = 16.00 g/mol
Molar mass of $\ce{O2}$ = 2 $\times$ 16.00 g/mol = 32.00 g/mol

Moles of $\ce{O2}$ = Mass of $\ce{O2}$ / Molar mass of $\ce{O2}$
Moles of $\ce{O2}$ = 0.27 g / 32.00 g/mol
Moles of $\ce{O2}$ = 0.0084375 mol

Rounding to two significant figures (due to 0.27 g), or three if we consider the initial masses, let's keep 3 sig figs for intermediate calculations and round at the end.
Moles of $\ce{O2}$ $\approx$ 0.00844 mol

The number of moles of oxygen that reacted is 0.00844\ mol.

The balanced chemical equation is: $\ce{2M(s) + O2(g) -> 2MO(s)}$
From the equation, 2 moles of M react with 1 mole of $\ce{O2}$.
Moles of M = Moles of $\ce{O2}$ $\times$ (2 moles M / 1 mole $\ce{O2}$)
Moles of M = 0.0084375 mol $\times$ 2
Moles of M = 0.016875 mol

Rounding to three significant figures:
Moles of M $\approx$ 0.0169 mol

The number of moles of metal (M) that reacted is 0.0169\ mol.

Molar mass of M = Mass of M / Moles of M
Molar mass of M = 1.00 g / 0.016875 mol
Molar mass of M $\approx$ 59.259 g/mol

Rounding to three significant figures:
Molar mass of M $\approx$ 59.3 g/mol

The molar mass of metal (M) is 59.3\ g/mol.

The calculated molar mass of metal (M) is approximately 59.3 g/mol.
Looking at the periodic table, the element with a molar mass closest to 59.3 g/mol is Cobalt (Co), which has a molar mass of approximately 58.93 g/mol.

The identity of the metal (M) is Cobalt (Co).