Answer
Astronaut's apparent weight: $58.8N$ away from the moon
With acceleration: $76.2N$ toward the moon
Work Step by Step
$F_G=G\frac{m_1m_2}{r^2}$
$F_{AW}=6.67\times10^{-11}\frac{Nm^2}{kg^2}\frac{75kg\times7.35\times10^{22}kg}{(2500\times10^3m)^2}=58.8N$ (away from the moon)
$\sum F=mg-N$
$N=mg_M-ma$
$N=58.8N-75kg\times (1.8\frac{m}{s^2})=-76.2N$ (toward the moon)