Verified answer

[tex]\displaystyle\bf\\1)\\\\Sin45^\circ-Cos60^\circ \cdot Cos90^\circ+Sin60^\circ=\frac{\sqrt{2} }{2} -\frac{1}{2} \cdot 0+\frac{\sqrt{3} }{2} =\\\\\\=\frac{\sqrt{2} }{2} +\frac{\sqrt{3} }{2} =\frac{\sqrt{2}+\sqrt{3} }{2} \\\\\\2)\\\\Sin30^\circ\cdot Cos30^\circ- Sin^{2} 45^\circ\cdot tg60^\circ=\frac{1}{2} \cdot\frac{\sqrt{3} }{2} -\Big(\frac{1}{\sqrt{2} } \Big)^{2}\cdot \sqrt{3} =[/tex]

[tex]\displaystyle\bf\\=\frac{\sqrt{3} }{4} -\frac{1}{2} \cdot \sqrt{3} =\frac{\sqrt{3} }{4} -\frac{\sqrt{3} }{2} =\frac{\sqrt{3} -2\sqrt{3} }{4} =-\frac{\sqrt{3} }{4} \\\\\\3)\\\\Cos30^\circ\cdot Sin60^\circ-tg45^\circ\cdot Ctg45^\circ\cdot Sin30^\circ=\frac{\sqrt{3} }{2} \cdot \frac{\sqrt{3} }{2} -1\cdot 1\cdot \frac{1}{2} =\\\\\\=\frac{3}{4} -\frac{1}{2} =\frac{1}{4}[/tex]