A variance (σ2) of 64 corresponds to a standard deviation (σ) of √64=8.
Mean μ=555. I'm guessing that 555 should be 55% or a score of 55.
(1) When X=60, Z=(60-μ)/σ for a large sample, that is, Z=(60-55)/8=⅝=0.625.
This Z-score corresponds to 0.734 approx or 73.40%. This means 73.40% of students at most score 60, implying that 26.6% have scores of 60 or more.
(2) and (3) When X=50, Z=(50-55)/8=-⅝=-0.625, corresponding to 26.6%. Therefore (2) 73.4% pass (above 50) while (3) 26.6% fail.