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(1)解：$3(2x - 1) = 15$
$2x - 1 = 5$
$2x = 6$
$x = 3$
(2)解：$\frac{x - 7}{3} - \frac{1 + x}{2} = 1$
$2(x - 7) - 3(1 + x) = 6$
$2x - 14 - 3 - 3x = 6$
$-x - 17 = 6$
$-x = 23$
$x = -23$

(1)不是
(2)解：因?yàn)辄c(diǎn)$C$是線段$AB$的“倍點(diǎn)”，$AB = 16cm$，所以有兩種情況：
情況一：當(dāng)$BC = 2AC$時(shí)，$AC + BC=AB$，即$AC + 2AC=16$，$3AC = 16$，解得$AC=\frac{16}{3}cm$；
情況二：當(dāng)$AC = 2BC$時(shí)，$AC + BC=AB$，即$2BC + BC=16$，$3BC = 16$，$BC=\frac{16}{3}cm$，所以$AC=2×\frac{16}{3}=\frac{32}{3}cm$。
綜上，$AC$的長(zhǎng)為$\frac{16}{3}cm$或$\frac{32}{3}cm$。

(1) $(-2)\otimes 3 = (-2)×3^{2} + 2×(-2)×3 + (-2)$
$= (-2)×9 + (-12) + (-2)$
$= -18 - 12 - 2$
$= -32$
(2) 由題意得：
$\frac{a + 1}{2}×3^{2} + 2×\frac{a + 1}{2}×3 + \frac{a + 1}{2} = 8$
$\frac{a + 1}{2}×9 + 3(a + 1) + \frac{a + 1}{2} = 8$
$\frac{9(a + 1) + 6(a + 1) + (a + 1)}{2} = 8$
$\frac{16(a + 1)}{2} = 8$
$8(a + 1) = 8$
$a + 1 = 1$
解得 $a = 0$
(3) 由題意得：
$m = 2\otimes x = 2x^{2} + 2×2×x + 2 = 2x^{2} + 4x + 2$
$n = (\frac{1}{4}x)\otimes 3 = \frac{1}{4}x×3^{2} + 2×\frac{1}{4}x×3 + \frac{1}{4}x$
$= \frac{1}{4}x×9 + \frac{3}{2}x + \frac{1}{4}x$
$= \frac{9x}{4} + \frac{6x}{4} + \frac{x}{4} = 4x$
$m - n = 2x^{2} + 4x + 2 - 4x = 2x^{2} + 2$
$\because x^{2} \geq 0$
$\therefore 2x^{2} + 2 \geq 2 ＞ 0$
$\therefore m ＞ n$