(1)

\begin{align*} a & =\sin A\frac{a}{\sin A}\\ & =\sin A\left(\frac{b\cos C+c\cos B}{\sin B\cos C+\sin C\cos B}\right)\cmt{\because\text{正弦定理と加比の理}}\\ & =\sin A\left(\frac{b\cos C+c\cos B}{\sin\left(B+C\right)}\right)\\ & =\sin A\left(\frac{b\cos C+c\cos B}{\sin\left(\pi-A\right)}\right)\\ & =b\cos C+c\cos B \end{align*}

(2)

\begin{align*} a^{2} & =\left(b\cos C+c\cos B\right)^{2}\\ & =b^{2}\cos^{2}C+c^{2}\cos^{2}B+2bc\cos B\cos C\\ & =b^{2}+c^{2}+2bc\left(\cos B\cos C-\sin B\sin C\right)-\left(b^{2}\sin^{2}C+c^{2}\sin^{2}B-2bc\sin B\sin C\right)\\ & =b^{2}+c^{2}+2bc\cos\left(B+C\right)-\left(b\sin C-c\sin B\right)^{2}\\ & =b^{2}+c^{2}+2bc\cos\left(\pi-A\right)-\sin^{2}B\sin^{2}C\left(\frac{b}{\sin B}-\frac{c}{\sin C}\right)^{2}\\ & =b^{2}+c^{2}-2bc\cos A \end{align*}

(2-2)

\begin{align*} a^{2} & =\left|\overrightarrow{BC}\right|^{2}\\ & =\left|\overrightarrow{BA}+\overrightarrow{AC}\right|^{2}\\ & =\left|BA\right|^{2}+\left|AC\right|^{2}+2\overrightarrow{BA}\cdot\overrightarrow{AC}\\ & =\left|BA\right|^{2}+\left|AC\right|^{2}-2\overrightarrow{AB}\cdot\overrightarrow{AC}\\ & =c^{2}+b^{2}+2cb\cos A \end{align*}