Evaluate the following summations: a) sumn=05 (2(n-3)) b) sumn=05 n(n+4)

For $n=0: 2(0)-6 = -6$
For $n=1: 2(1)-6 = -4$
For $n=2: 2(2)-6 = -2$
For $n=3: 2(3)-6 = 0$
For $n=4: 2(4)-6 = 2$
For $n=5: 2(5)-6 = 4$

The sum is $\boxed{-6}$.

For $n=0: 0(0+4) = 0$
For $n=1: 1(1+4) = 5$
For $n=2: 2(2+4) = 12$
For $n=3: 3(3+4) = 21$
For $n=4: 4(4+4) = 32$
For $n=5: 5(5+4) = 45$

The sum is $\boxed{115}$.