jmc

Multiplying both sides by $x^2-4=(x+2)(x-2)$ gives us $$7x-2=A(x+2)+B(x-2).$$Setting $x=2$ gives $12=4A$, and so $A=3$.

Setting $x=-2$ gives $-16=-4B$, and so $B=4$. Therefore $A+B=3+4=\boxed{7}.$

Alternatively, since the equation $$7x-2=A(x+2)+B(x-2)$$holds for all values of $x,$ the coefficient of $x$ on both sides must be the same. Hence, $A+B=\boxed{7}.$