21. Solution: (1) When $a = 0$, it clearly does not satisfy the problem conditions, so $a \neq 0$. & 1 mark \\
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$f'(x) = 3ax^2 - 2x$. Setting $f'(x) = 0$, we get $x = 0$ or $x = \frac{2}{3a}$. & $f'(x) = 3ax^2 - 2x$. Setting $f'(x) = 0$, we get $x = 0$ or $x = \frac{2}{3a}$. \\
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From the problem conditions, $1 < \frac{2}{3a} < 3$. Solving, we get $\frac{2}{9} < a < \frac{2}{3}$, i.e., the range of $a$ is $\left(\frac{2}{9}, \frac{2}{3}\right)$. & 4 marks \\
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(2) When $a = 0$, $f(x) = -x^2$ has minimum value $f(2) = -4$ on $[-1, 2]$. & 5 marks \\
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Since $x \in [-1, 2]$, $\therefore 3x - \frac{2}{a} \leq 0$. & When $0 < a \leq \frac{1}{3}$, $\frac{2}{a} \geq 6$. $f'(x) = ax\left(3x - \frac{2}{a}\right)$. \\
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6 marks &  \\
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$\because f(2) - f(-1) = (8a - 4) - (-a - 1) = 9a - 3 \leq 0$, $\therefore f(x)_{\min} = f(2) = 8a - 4$. & 7 marks \\
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When $a > \frac{1}{3}$, $f'(x) = ax\left(3x - \frac{2}{a}\right)$, $0 < \frac{2}{3a} < 2$. When $x \in [-1, 0) \cup \left(\frac{2}{3a}, 2\right]$, $f'(x) > 0$; &  \\
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When $x \in \left(0, \frac{2}{3a}\right)$, $f'(x) < 0$. & 8 marks \\
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$\therefore f(x)_{\min} = \min\left\{f(-1), f\left(\frac{2}{3a}\right)\right\}$. &  \\
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9 marks &  \\
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Since $a > \frac{1}{3}$, $\therefore 27a^3 + 27a^2 - 4 > 0$, $\frac{27a^3 + 27a^2 - 4}{27a^2} > 0$. & 10 marks \\
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$\therefore f(x)_{\min} = f(-1) = -a - 1$. & 11 marks \\
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In summary, when $0 \leq a \leq \frac{1}{3}$, $f(x)_{\min} = 8a - 4$; when $a > \frac{1}{3}$, $f(x)_{\min} = -a - 1$. & 12 marks \\
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1 mark &  \\
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