Verified answer

[tex]\displaystyle\bf\\1)\\\\6Cos105^\circ Cos15^\circ=6Cos(90^\circ+15^\circ) Cos15^\circ=\\\\\\=6\cdot(-Sin15^\circ)\cdot Cos15^\circ=-3\cdot(2Sin15^\circ Cos15^\circ)=\\\\\\=-3Sin30^\circ=-3\cdot\frac{1}{2} =-1,5\\\\2)\\\\\frac{Sin13^\circ Cos47^\circ+Sin47^\circ Cos13^\circ}{Cos98^\circ Cos38^\circ +Sin98^\circ Sin38^\circ} =\frac{Sin(13^\circ+47^\circ)}{Cos(98^\circ-38^\circ)} =\\\\\\=\frac{Sin60^\circ}{Cos60^\circ} =tg60^\circ=\frac{\sqrt{3} }{3}[/tex]