Question

A point charge q = +40.0 µC moves from A to B separated by a distance d = 0.180 m in the presence of an external electric field of magnitude 275 N/C directed toward the right.

Answer 1

a) You got the right formula but the units are wrong!
F = qE = (40x10^-6) x 275 = 1.10x10^-2 N Note, 3 significant figures should be used.
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b) You got the right answer but forgot units.
The easiest method is: work done = force x displacement.

If A is on the left then W = 0.0110 x 0.180 = 1.98x10^-3 J
If A is on the right then W = -0.0110 x 0.180 = -1.98x10^-3 J
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c) The change in potential energy is simply minus whatever the answer to part b) is.

When a field, for example accelerates, a charge, the charge loses potential energy (it gets converted to kinetic energy).

(Just think of gravity. When something falls, the loss of potential energy = gain in kinetic energy; the lost potential energy is the work done by the gravitational field.)
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d) The potential difference (V) is defined as energy transferred/charge

If A is on the left V = -1.98x10^-3/(40x10^-6) = -49.5V
If A is on the right V = 1.98x10^-3/(40x10^-6) = 49.5V

The + or - signs are potentially confusing - it depends if they want the potential of A relative to B, or of B relative to A.