jmc

Because the polynomial has rational coefficients, the radical conjugate of each of the given roots must also be roots of the polynomial. However, $5$ is rational, so the argument does not apply to it; furthermore, $1-\sqrt{2}$ and $1+\sqrt{2}$ are each other's radical conjugates, so the only other root that the polynomial must have is $3-\sqrt{7}.$ This makes at least $1+4=5$ roots.

Furthermore, the polynomial $$(x-1+\sqrt{2})(x-1-\sqrt{2})(x-3+\sqrt{7})(x-3-\sqrt{7})(x-5)=(x^2-2x-1)(x^2-6x+2)(x-5)$$has roots $1\pm \sqrt{2},$ $3\pm \sqrt{7},$ and 5, and has rational coefficients. Hence, $\boxed{5}$ is the smallest possible degree of the polynomial.