在△ABC中角ABC的对边分别是abc满足bsinB+csinC=sinA·a-2bsinC1求角A的余弦值；2若D是边AB的中点且CD=2求⁤b+ sqrt 2c⁤的取值范围

(1)将已知条件改写为$\frac{b}{a}\cdot\frac{\sin B}{\sin A}+\frac{c}{a}\cdot\frac{\sin C}{\sin A}=1-2\frac{b}{a}\cdot\sin C$，再利用正弦定理将$\frac{b}{a}$和$\frac{c}{a}$表示为$\sin B$和$\sin C$的函数，得到$$\frac{\sin B}{\sin A}\cdot\frac{\sin^2 C}{\sin B\sin C}+\frac{\sin C}{\sin A}\cdot\frac{\sin^2 B}{\sin B\sin C}=1-2\frac{\sin B}{\sin A}\cdot\sin C$$$$\Rightarrow\frac{\sin C}{\sin A}\sin B+\frac{\sin B}{\sin A}\sin C=2\sin B\sin C-\sin A\sin B$$$$\Rightarrow\sin(A+B)\sin C=\sin A\sin B(2\cos C-1)$$$$\Rightarrow\sin A=\frac{\sin B\sin C}{\sin(A+B)\cos C-\cos(A+B)\sin C}$$由余弦定理可知$\cos A=\frac{b^2+c^2-a^2}{2bc}$，代入上式化简得$$\sin A=\frac{2bc\sin B\sin C}{b^2+c^2-a^2+2bc\cos A}=\frac{2bc\sin B\sin C}{(b+c)^2-a^2}$$$$\Rightarrow\cos A=\frac{(b+c)^2-a^2-4b^2\sin^2 C-4c^2\sin^2 B}{2bc}$$

(2)根据余弦定理和中线定理，有$$b^2+c^2-2bc\cos A=2\left(\frac{a^2}{2}-\frac{b^2}{4}\right)+2\left(\frac{a^2}{2}-\frac{c^2}{4}\right)$$$$\Rightarrow b^2+c^2-2bc\cos A=a^2-b^2-c^2$$$$\Rightarrow b^2+c^2-2bc\cos A=2bc\cos A-2b^2\sin^2 C-2c^2\sin^2 B$$$$\Rightarrow b^2+2bc\cos A+c^2=2bc\cos A+2b^2\sin^2 C+2c^2\sin^2 B$$$$\Rightarrow (b+c\cos A)^2=b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C+2$$$$\Rightarrow b+\sqrt{2}c\cos A\in[\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C},\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C+2}]$$注意到$b\sin C+c\sin B\leq b+c$，因此$$b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C\leq (b\sin C+c\sin B)^2\leq (b+c)^2$$$$\Rightarrow \sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C}\leq b+c$$$$\Rightarrow b+\sqrt{2}c\cos A\in[\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C},b+c\sqrt{2}]$