Verified answer

[tex]\displaystyle\bf\\1)\\\\\frac{2c^{2} }{c-8} -2c=\frac{2c^{2} -2c\cdot(c-8)}{c-8} =\frac{2c^{2}-2c^{2} +16c }{c-8} =\frac{16c}{c-8} \\\\\\2)\\\\\frac{4}{a-c} -\frac{2}{a+c} -\frac{4c}{a^{2} -c^{2} } =\frac{4}{a-c} -\frac{2}{a+c} -\frac{4c}{(a -c)(a+c)} =\\\\\\=\frac{4\cdot(a+c)-2\cdot(a-c)-4c}{(a-c)(a+c)} =\frac{4a+4c-2a+2c-4c}{(a-c)(a+c)} =\\\\\\=\frac{2a+2c}{(a-c)(a+c)} =\frac{2(a+c)}{(a-c)(a+c)} =\frac{2}{a-c}[/tex]