5. In the Cartesian coordinate system, regardless of the value of $m$, the parabola $y=m x^{2}+(2 m+1) x-(3 m+2)$ does not pass through the points on the line $y=-x+1$. The coordinates of all points that meet this condition are (write down the coordinates of all such points).

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5. $(1,0),(-3,4),\left(\frac{3}{2},-\frac{1}{2}\right)$.

From $y=m x^{2}+(2 m+1) x-(3 m+2)$
$$
=m(x+3)(x-1)+(x-2)
$$

we can see that the parabola must pass through the points $A(1,-1)$ and $B(-3,-5)$.
Draw lines parallel to the $y$-axis through points $A$ and $B$ to intersect the line $y=-x+1$ at points $C(1,0)$ and $D(-3,4)$.

The line through points $A$ and $B$ intersects $y=-x+1$ at point $E\left(\frac{3}{2},-\frac{1}{2}\right)$.
Thus, points $C$, $D$, and $E$ satisfy the conditions.