How do I do this limiting reactants problem about magnesium oxide?

Number 1 is not correct, always use the second weighing mass the sample of MgO had to be cooled before weighing, I suggest reweighing it again and again until the mass reads the same.

Addressing number 2:
Write a balanced equation
#2Mg+O_2->2MgO#

2)
Determine the limiting reagent:
#2.40 cancel("g Mg")* (1 cancel("mol Mg"))/ (24.3 cancel("g Mg")) *(1 cancel("mol O"_2))/ (2 cancel("mol Mg"))* (32 "g O"_2)/ (1 cancel("mol O"_2))= 1.58 g O_2# required to react, since we have that amount, #O_2# is in excess and Mg is said to be limiting

3)
Stoichometry using limiting reagent:
# 2.40 cancel("g Mg")* (1 cancel("mol Mg"))/ (24.3 cancel("g Mg"))* (2 cancel("mol MgO"))/ (2 cancel("mol Mg"))* (40.3"g MgO")/ (1 cancel("mol MgO"))= 3.98 g MgO#

4) Percent Error
#(%"Error") = ("Actual - Theoretical")/("Theoretical") xx 100%#

#(%"Error") = |(3.26 - 3.98)/(3.98)| xx 100% = 18.1%#

Percent yield (optional):
#("Experimental yield")/("Theoretical yield")*100#
#(3.26 g)/ (3.98 g)*100= 81.9%#

Note: Your experimental yield should never exceed your theoretical yield or there is problem with your lab methodology