De Moivre's theorem uses this general formula z = r(cos α + i sin α) that is where we can have the form a + bi. If the given is raised to a certain number, then the r is raised to the same number while the angles are being multiplied by that number.
For 1) [3cos(27))+isin(27)]^5 we first apply the concept I mentioned above where it becomes
[3^5cos(27*5))+isin(27*5)] and then after simplifying we get, [243 (cos (135) + isin (135))]
it is then further simplified to 243 (-1/ √2) + 243i (1/√2) = -243/√2 + 243/√2 i
and that is the answer.
For 2) [2(cos(40))+isin(40)]^6, we apply the same steps in 1)
[2^6(cos(40*6))+isin(40*6)],
[64(cos(240))+isin(240)] = 64 (-1/2) + 64i (-√3 /2)
And the answer is -32 -32 √3 i
Summary:
1) -243/√2 + 243/√2 i
2)-32 -32 √3 i