a) Gọi I(x; y; z)

Ta có: I là trọng tâm tam giác ABC nên \(\left\{ \begin{array}{l}x = \frac{3}\\y = \frac{3}\\z = \frac{{ - 3 + 5 + ( - 5)}}{3}\end{array} \right.\) ⇔ \(\left\{ \begin{array}{l}x = 3\\y = 4\\z =  - 1\end{array} \right.\).

Vậy I(3; 4; −1).

b) Gọi G(x0; y0; z0), theo đề ta có: DG = 3IG nên DG = \(\frac{3}{4}\)DI suy ra \[\overrightarrow {DG}  = \frac{3}{4}\overrightarrow {DI} \].

Suy ra \(\left\{ \begin{array}{l}{x_0} + 1 = \frac{3}{4}.4\\{y_0} + 4 = \frac{3}{4}.8\\{z_0} - 3 = \frac{3}{4}.( - 4)\end{array} \right.\) ⇒ \(\left\{ \begin{array}{l}{x_0} = 2\\{y_0} = 2\\{z_0} = 0\end{array} \right.\).

Vậy G(2; 2; 0).