jmc

Since $x^2+4x+3=(x+1)(x+3)$ has degree $2$, the remainder must be of the form $ax+b$ for some constants $a$ and $b$. Let $q(x)$ be the quotient of the division, so $$2x^6-x^4+4x^2-7=(x+1)(x+3)q(x)+ax+b.$$Plugging in $x=-1$ gives us: $$2(-1{)}^6-(-1{)}^4+4(-1{)}^2-7=0+a(-1)+b,$$which simplifies to $$b-a=-2.$$Plugging in $x=-3$ gives us: $$2(-3{)}^6-(-3{)}^4+4(-3{)}^2-7=0+a(-3)+b,$$which simplifies to $$b-3a=1406.$$Solving this system of equations gives us $a=-704$ and $b=-706$, and so our remainder is $\boxed{-704x-706}$.