Answer
$F_1=\frac{1}{2}m(a_1+a_2)$
$F_2=\frac{1}{2}m(a_1-a_2)$
Work Step by Step
The total force on the object when pushed by two persons in the same direction is
$F_1+F_2=ma_1$....eq(1)
The total force on the object when pushed by two persons in the opposite direction is
$F_1-F_2=ma_2$.....eq(1)
Adding eq (1)and eq(2), we obtain:
$2F_1=m(a_1+a_2)$
$\implies F_1=\frac{1}{2}m(a_1+a_2)$
Now, eq(1):
$F_2=ma_1-F_1$
We plug in $ F_1=\frac{1}{2}m(a_1+a_2)$ to obtain:
$F_2=ma_1-\frac{1}{2}m(a_1+a_2)$
$\implies F_2=\frac{2ma_1-m(a_1+a_2)}{2}$
$\implies F_2=\frac{1}{2}m(a_1-a_2)$
