Verified answer

[tex]\displaystyle 1)\frac{16-y^2}{4-y}=\frac{(4-y)(4+y)}{4-y}=4+y\\ \\2)1-x^2-2xp-p^2=1-(x^2+2xp+p^2)=1-(x+p)^2=(1-(x+p))(1+(x+p))=\\(1-x-p)(1+x+p)\\\\3)\frac{121m^2-1}{121m^2-22m+1}=\frac{(11m-1)(11m+1)}{(11m-1)^2}=\frac{11m+1}{11m-1}[/tex]

[tex]\displaystyle\bf\\1)\\\\\frac{16-y^{2} }{4-y} =\frac{4^{2} -y^{2} }{4-y} =\frac{(4-y)(4+y)}{4-y} =\boxed{4+y}\\\\\\2)\\\\1-x^{2} -2xp-p^{2} =1-(x^{2} +2xp+p^{2} )=1^{2} -(x+p)^{2} =\\\\=\boxed{(1-x-p)(1+x+p)}\\\\\\3)\\\\\frac{121m^{2} -1}{121m^{2} -22m+1} =\frac{(11m)^{2} -1^{2} }{(11m)^{2}-2\cdot 11m+ 1^{2} } =\\\\\\=\frac{(11m-1)(11m+1)}{(11m-1)^{2} } =\boxed{\frac{11m+1}{11m-1} }[/tex]