jmc

The remainder when $p(x)$ is divided by $(x+1)(x+5)$ is of the form $ax+b.$ Thus, we can let $$p(x)=(x+1)(x+5)q(x)+ax+b,$$where $q(x)$ is the quotient in the division.

By the Remainder Theorem, $p(-1)=5$ and $p(-5)=-7.$ Setting $x=-1$ and $x=-5$ in the equation above, we get $$\begin{array}{rl}-a+b& =5,\\ -5a+b& =-7.\end{array}$$Solving, we find $a=3$ and $b=8,$ so the remainder is $\boxed{3x+8}.$