Solution:
(1) Since $S_n = n^2 - n$,

Let $n=1$, we get $a_1 = 0$

$a_n = S_n - S_{n-1} = 2(n-1)$, for $(n \geq 2)$

Therefore, $a_n = 2(n-1)$.

Since sequence $\{b_n\}$ is a geometric sequence, $b_2 = a_2 = 2$, $b_4 = a_5 = 8$,

Therefore, $\frac{b_4}{b_2} = q^2 = 4$,

And since all terms are positive,

$q = 2$, therefore $b_n = 2^{n-1}$.

(2) From (1), we get: $c_n = (n-1) \cdot 2^n$,

Therefore, $T_n = 0 + (2-1) \cdot 2^2 + (3-1) \cdot 2^3 + \cdots + (n-1) \cdot 2^n = 1 \cdot 2^2 + 2 \cdot 2^3 + \cdots + (n-1) \cdot 2^n$,

$2T_n = 1 \cdot 2^3 + 2 \cdot 2^4 + \cdots + (n-2) \cdot 2^n + (n-1) \cdot 2^{n+1}$,

$-T_n = 2^2 + 2^3 + 2^4 + \cdots + 2^n - (n-1) \cdot 2^{n+1} = \frac{2^2(1-2^{n-1})}{1-2} - (n-1) \cdot 2^{n+1} = 2^{n+1} - (n-1) \cdot 2^{n+1} - 4$,

Therefore, $T_n = (n-2) \cdot 2^{n+1} + 4$.

Thus, the final answers are:
(1) $a_n = 2(n-1)$, $b_n = 2^{n-1}$
(2) $T_n = (n-2) \cdot 2^{n+1} + 4$
So, the boxed answers are:
(1) $\boxed{a_n = 2(n-1), b_n = 2^{n-1}}$
(2) $\boxed{T_n = (n-2) \cdot 2^{n+1} + 4}$