1)
[tex3]
\begin{align}
2\cos(2\alpha)+3\cos(\alpha)+1=0&\implies 4\cos^2\!\!\alpha+3\cos\alpha-1=0\\
&\implies \left\{\begin{array}{ll}\cos\alpha=\dfrac{-3-\sqrt{9+16}}{8}=-1\quad\text{impossível já que }\alpha<\pi\\\text{ou}\\\cos\alpha=\dfrac{-3+5}{8}=-\dfrac{1}{4}\end{array}\right.\\
&\implies \left\{\begin{array}{ll}\sin\alpha=\sqrt{1-\cos^2\alpha}=\dfrac{\sqrt{15}}{4}\\\text{ou}\\\sin\alpha=-\sqrt{1-\cos^2\alpha}=-\dfrac{\sqrt{15}}{4}\quad\text{impossível já que }0<\alpha<\pi\end{array}\right.\\
&\implies \sin\alpha=\frac{\sqrt{15}}{4}
\end{align}\\[96pt]
[/tex3]
2)
[tex3]
\begin{align}
\text{Temos }\left.\begin{array}[m]{rl}
AB\sin\widehat{B}=AC\sin\widehat{C}\\
AC\cos\widehat{C}+AB\cos\widehat{B}=BC
\end{array}\right\}
&\implies AC\cos\widehat{C}+AC\dfrac{\sin\widehat{C}}{\sin\widehat{B}}\cos\widehat{B}=BC\\
&\implies AC=\dfrac{BC}{\cos\widehat{C}+\cos\widehat{B}\cdot\dfrac{\sin\widehat{C}}{\sin\widehat{B}}}\\
&\implies AC=\dfrac{BC}{\cos\alpha+\cos(\dfrac{\alpha}{2})\cdot \dfrac{\sin\alpha}{\sin(\dfrac{\alpha}{2})}}\\
&\implies AC=\dfrac{BC}{2\cos^2(\dfrac{\alpha}{2})-1+\cos(\dfrac{\alpha}{2})\cdot \dfrac{2\sin(\dfrac{\alpha}{2})\cos(\dfrac{\alpha}{2})}{\sin(\dfrac{\alpha}{2})}}\\
&\implies AC=\dfrac{BC}{4\cos^2(\dfrac{\alpha}{2})-1}\\
&\implies AC=\dfrac{BC}{2(2\cos^2(\dfrac{\alpha}{2})-1)+1}=\dfrac{BC}{2\cos\alpha+1}\\
&\implies AC=\dfrac{\sqrt{15}}{5}\cdot \dfrac{2}{3}=2\,\dfrac{\sqrt{15}}{15}
\end{align}\\[48pt]
[/tex3]
[tex3]
\boxed{\\\hspace{1cm}\\\hspace{1cm}AC=2\,\dfrac{\sqrt{15}}{15}\hspace{1cm}\\}
[/tex3]