jmc

First of all, $6$ has a remainder of $1$ when divided by $5,$ thus any power of $6$ will have a remainder of $1$ when divided by $5.$

As for $7,$ it has a remainder of $2$ when divided by $5$, so let us look at the powers of $2$: $$\begin{array}{rl}{2}^1& \equiv 2(\mathrm{mod}5)\\ 2^2& \equiv 4(\mathrm{mod}5)\\ 2^3& \equiv 3(\mathrm{mod}5)\\ 2^4& \equiv 1(\mathrm{mod}5).\end{array}$$Since $2^4\equiv 1(\mathrm{mod}5),$ we see that $2^7\equiv 2^3\cdot 2^4\equiv 3(\mathrm{mod}5),$ hence the remainder of $7^4$ when divided by $5$ is $3.$

Now, there is a bit of a shortcut we can use for $8.$ Since $8\equiv -2(\mathrm{mod}5),$ we can see that $8^6\equiv (-2{)}^6\equiv 2^6\equiv 2^2\cdot 2^4\equiv 4(\mathrm{mod}5),$ hence our desired remainder is $4.$

Adding them up, we have $4+3+1\equiv \boxed{3}(\mathrm{mod}5).$