Ok, your drawing is not easy to work with since d1 < d2 which is not as it is in your drawings
In both figures, I will name the parrallelogram ABCD (clockwise, A at the bottom left) with AC the diagonal D1 and BD the diagonal D2. Fianlly, E is the intersection of d1 and d2.
You know that d1 and d2 intersect at their middle. So m(AE)=m(EC) and m(BE)=m(ED).
Let's beggin with the second parallelogram. We got m(AB)=m(BC)=m(CD)=m(DA). So in particular, the triangle ABD is isosceles with m(AB)=m(DA). And since AE cuts the side BD at its middle so it is also perpendicular to BD (because ABD is isosceles). So, the triangle ABE is right-angled triangle. We have m(AE)=7 and m(BE)=24 so m(AB)=sqrt(49+24^2) = 25. Let's find the measure of the angle BAE so sin(BAE)=24/25 so m(BAE)=73.74 degrees. Since AE bisects the angle BAD so m(BAE)=m(EAD)=73.74 degrees.
Ok, I'll name the foot of your height F (So the height is BF). So the angle BAF measures 180-2*73.74 = 32.52 degrees. Now, we are almost done. The triangle BAF is right-angled so sin(BAF) = m(BF)/25 so sin(32.52)=m(BF)/25 so m(BF)=13.44. Done !
For the first one, since you have a and b, it is easy. You don't need the angles. Heron formula for the area of a triangle is sufficient. Use it 1 time : to find the area of the triangle ABD. So ABD has sides 74, sqrt(937) and 51 so the p=semiperimeter=(74+sqrt(937)+51)/2=77.805
Area(ABD)=sqrt(p*(p-74)*(p-sqrt(937))*(p-51))=612. And since the triangle BDC has the same length of sides so its area is also 612 and so the area of the parallelogram is Area(BDC)+Area(ABD)=2*612=1224.
If you don't understand how to find a and b just reply and ask for that.