已知$\dfrac{1+\tan (θ+720^{\circ})}{1-\tan (θ-360^{\circ})}=3+2 \sqrt{2}$，求：$[\cos ^{2}(π- θ)+\sin (π+θ)·\cos (π-θ)+2\sin ^{2}(θ-π)]· \dfrac{1}{\cos ^{2}(-θ-2π)}$的值．

1. 利用诱导公式化简已知等式

由三角函数的诱导公式，可得 $\tan(\theta+720^{\circ}) = \tan(\theta+2\times 360^{\circ}) = \tan\theta$，$\tan(\theta-360^{\circ}) = \tan\theta$．
代入原方程可得：
$\dfrac{1+\tan\theta}{1-\tan\theta} = 3+2\sqrt{2}$

2. 解方程求出 $\tan\theta$ 的值

由等式 $\dfrac{1+\tan\theta}{1-\tan\theta} = 3+2\sqrt{2}$ 去分母得：
$1+\tan\theta = (3+2\sqrt{2})(1-\tan\theta)$
整理得：
$(4+2\sqrt{2})\tan\theta = 2+2\sqrt{2}$
从而解得：
$\tan\theta = \dfrac{2+2\sqrt{2}}{4+2\sqrt{2}} = \dfrac{2(1+\sqrt{2})}{2(2+\sqrt{2})} = \dfrac{1+\sqrt{2}}{\sqrt{2}(\sqrt{2}+1)} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$

3. 利用诱导公式化简待求表达式

对于待求表达式：
$E = [\cos^{2}(\pi-\theta)+\sin(\pi+\theta)\cdot\cos(\pi-\theta)+2\sin^{2}(\theta-\pi)]\cdot \dfrac{1}{\cos^{2}(-\theta-2\pi)}$
根据诱导公式化简各项：

• $\cos(\pi-\theta) = -\cos\theta \implies \cos^2(\pi-\theta) = \cos^2\theta$
• $\sin(\pi+\theta) = -\sin\theta$
• $\cos(\pi-\theta) = -\cos\theta$
• $\sin(\theta-\pi) = -\sin(\pi-\theta) = -\sin\theta \implies \sin^2(\theta-\pi) = \sin^2\theta$
• $\cos(-\theta-2\pi) = \cos\theta \implies \cos^2(-\theta-2\pi) = \cos^2\theta$

将上述化简结果代入待求式得：
$E = (\cos^2\theta + \sin\theta\cos\theta + 2\sin^2\theta) \cdot \dfrac{1}{\cos^2\theta}$

4. 将表达式用 $\tan\theta$ 表示并求值

展开并化简待求式：
$E = \dfrac{\cos^2\theta}{\cos^2\theta} + \dfrac{\sin\theta\cos\theta}{\cos^2\theta} + 2\cdot\dfrac{\sin^2\theta}{\cos^2\theta}$
$E = 1 + \tan\theta + 2\tan^2\theta$
将 $\tan\theta = \dfrac{\sqrt{2}}{2}$ 代入，得：
$E = 1 + \dfrac{\sqrt{2}}{2} + 2 \times \left(\dfrac{\sqrt{2}}{2}\right)^2 = 1 + \dfrac{\sqrt{2}}{2} + 1 = 2 + \dfrac{\sqrt{2}}{2}$