[tex]\displaystyle\bf\\1)\\\\\Big(2b+\frac{1}{4} c\Big)\Big(\frac{1}{4} c-2b\Big)=\Big(\frac{1}{4} c\Big)^{2} -\Big(2b\Big)^{2} =\frac{1}{16} c^{2} -4b^{2} \\\\\\2)\\\\\Big(\frac{4}{5}a -\frac{1}{4} c\Big)\Big(\frac{1}{4} c+\frac{4}{5} a\Big)=\Big(\frac{4}{5} a\Big)^{2} -\Big(\frac{1}{4} c\Big)^{2} =\frac{16}{25}a^{2} -\frac{1}{16} c^{2}[/tex]