$\frac{a-1}{2a}\ast (\frac{a+3}{a+1}-\frac{a^2-5}{a^2-1})=\frac{a-1}{2a}+(\frac{a+3}{a+1}-\frac{a^2-5}{(a-1)(a+1)})=\frac{a-1}{2a}\ast \frac{(a+3)(a-1)-(a^2-5)}{(a+1)(a-1)}=\frac{a-1}{2a}\ast \frac{a^2+3a-a-3-a^2+5}{(a+1)(a-1)}=\frac{2a+2}{2a(a+1)}=\frac{1}{a}$

$(\frac{c+3}{c-3}-\frac{c}{c+3})\ast \frac{c-3}{c+1}=\frac{(c+3)(c+3)-c(c-3)}{(c-3)(c+3)}\ast \frac{c-3}{c+1}=\frac{c^2+3c+3c+9-c^2+3c}{(c+3)(c+1)}=\frac{9c+9}{(c+3)(c+1)}=\frac{9(c+1)}{(c+3)(c+1)}=\frac{9}{c+3}$

$(\frac{14+a^2}{a^2-4}-\frac{a-4}{a+2})\ast \frac{a-2}{6}=(\frac{14+a^2}{(a-2)(a+2)}-\frac{a-4}{a+2})\ast \frac{a-2}{6}=\frac{14+a^2-(a-4)(a-2)}{(a-2)(a+2)}\ast \frac{a-2}{6}=\frac{14+a^2-a^2+4a+2a-8}{(a-2)(a+2)}\ast \frac{a-2}{6}=\frac{6a+6}{6(a+2)}=\frac{6(a+1)}{6(a+2)}=\frac{a+1}{a+2}$

$(\frac{a}{a-4}-\frac{a-4}{a+4})\ast \frac{a+4}{4}=(\frac{a(a+4)-(a-4)(a-4)}{(a-4)(a+4)})\ast \frac{a+4}{4}=\frac{a^2+4a-a^2+8a-16}{4(a-4)}=\frac{12a-16}{4(a-4)}=\frac{4(3a-4)}{4(a-4)}=\frac{3a-4}{a-4}$

$(\frac{y+1}{y-1}-\frac{y-1}{y+1})\ast \frac{y+1}{4y}=\frac{(y+1)(y+1)-(y-1)(y-1)}{(y-1)(y+1)}\ast \frac{y+1}{4y}=\frac{y^2+y+y+1-(y^2-y-y+1)}{4y(y-1)}=\frac{y^2+y+y+1-y^2+y+y-1}{4y(y-1)}=\frac{4y}{4y(y-1)}=\frac{1}{y-1}$

$(\frac{1+a}{1-a}-\frac{1-a}{1+a}):\frac{2a}{1-a}=\frac{(1+a)(1+a)-(1-a)(1-a)}{(1-a)(1+a)}\ast \frac{1-a}{2a}=\frac{1+a+a+a^2-(1-a-a+a^2)}{1+a}\ast \frac{1}{2a}=\frac{1+a+a+a^2-1+a+a-a^2}{1+a}\ast \frac{1}{2a}=\frac{4a}{(1+a)2a}=\frac{2}{1+a}$

$\frac{4y}{y-1}\ast (\frac{y}{8}-\frac{1}{4}+\frac{1}{8y})=\frac{4y}{y-1}\ast \frac{y\ast y-1\ast 2y+1}{8y}=\frac{4y}{y-1}\ast \frac{y^2-2y+1}{8y}=\frac{(y-1{)}^2}{2(y-1)}=\frac{y-1}{2}$

$(\frac{a}{8}+\frac{1}{3}+\frac{1}{6a}):\frac{a+1}{12a}=\frac{a\ast 3a+1\ast 8a+1\ast 4}{24a}\ast \frac{12a}{a+1}=\frac{3a^2+8a+4}{2(a+1)}$