71. (1) $a^{2}+a+1 \geqslant \frac{3}{4}(a+1)^{2} \Leftrightarrow (a-1)^{2} \geqslant 0$, so,
$$a+\frac{1}{a+1} \geqslant \frac{3}{4}(a+1)$$

Similarly,
$$\begin{array}{l} b+\frac{1}{b+1} \geqslant \frac{3}{4}(b+1) \\ c+\frac{1}{c+1} \geqslant \frac{3}{4}(c+1) \end{array}$$

Multiplying these inequalities, we get
$$\begin{array}{l} \left(a+\frac{1}{a+1}\right)\left(b+\frac{1}{b+1}\right)\left(c+\frac{1}{c+1}\right) \geqslant \\ \frac{27}{64}(a+1)(b+1)(c+1) \geqslant \\ \frac{27}{64} \cdot 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}= \\ \frac{27}{8} \sqrt{a b c} \geqslant \frac{27}{8} \end{array}$$
(2) $a^{2}+1 \geqslant \frac{2}{3}\left(a^{2}+a+1\right) \Leftrightarrow (a-1)^{2} \geqslant 0$, so,
$$\begin{array}{l} 27\left(a^{3}+a^{2}+a+1\right)\left(b^{3}+b^{2}+b+1\right)\left(c^{3}+c^{2}+c+1\right)= \\ 27\left(a^{2}+1\right)(a+1)\left(b^{2}+1\right)(b+1)\left(c^{2}+1\right)(c+1) \geqslant \\ 27 \cdot \frac{2}{3}\left(a^{2}+a+1\right) \cdot(a+1) \cdot \frac{2}{3}\left(b^{2}+b+1\right) \cdot(b+1) \cdot \\ \frac{2}{3}\left(c^{2}+c+1\right) \cdot(c+1) \cdot 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}= \\ 64\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right) \sqrt{a b c} \geqslant \\ 64\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)= \\ 8\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right)(a+1)(b+1)(c+1) \geqslant \\ 8\left(a^{2}+a+1\right)\left(b^{2}+b+1\right)\left(c^{2}+c+1\right) \end{array}$$