Find the area of the parallelogram whose vertices are at A(0,8), B(-2,6), C(-4,-6) and D(-2,-4).

Hello!
This figure is really a parallelogram, for example because the opposite sides have the same length: |AB| = |CD| = 2sqrt(2) and |BC| = |AD| = 2sqrt(37). Or we can check that the opposite sides have the same slope.
The area of a parallelogram ABCD is twice the area of the triangle ABC (or BCD,  or CDA, or DAB ). Therefore it is |AB|*|BC|*|sin(B)|.
The simplest way to compute this for the points with known coordinates is to note that this expression is the absolute value of the cross product:
A = |vec(BA) xx vec(BC)| = |lt2,2gt xx lt-2,-12gt| =
= |2*(-12)-2*(-2)| = |-24 + 4| = |-20| = 20.
This is the answer. If you don't know the cross product, you can use Heron's formula for any mentioned triangle (and multiply by 2).