Two boxes connected by a lightweight cord are resting on a table. The
boxes have masses of 12 kg and 10 kg. A horizontal force FP of 40 N is
applied by a person to the 10 kg box as shown in the figure. Find:

a.the acceleration of each box; and


b.the tension in the cord.

Hi again pj4444!!

In this problem I will assume that the force FP is applied pulling the
10kg box, and in a direction such the 10kg box (Box 1)pull the 12kg
box (Box 2):

   __________ T    T __________
   |        |-->  <--|        |
   |  12kg  |--------| 10kg   |-----> FP = 40N
   ----------        ----------


Because the two boxes are vinculated they will have the same
acceleration.
If A is the acceleration of the system and T is the tension of the
cord, we have the following equations for each box:

Box 1:
FP - T = m1 * A  
It means that the resultant force of FP and T is applied to the Box 1.

Box 2:
T = m2 * A
Only T is applied on the Box 2.

If we add these equations we have:
FP - T + T = m1 * A + m2 * A
then
FP = (m1 + m2) * A
then 
A = FP / (m1 + m2) = 40 / (10 + 12) = 40/22 = 1.818 m/s^2 .

This solve the part a).

We had for the Box 2:
T = m2 * A
then
T = 12 * 1.818 = 21.816 N .

This solve part b).

I hope this helps you, but if you need a clarification, please post a
request for it.

Best regards.
livioflores-ga