The equation of a curve is $y = (m+1)x^2 - 8x + 3m$, where $m$ is a constant.

1. Substitute $m = 1$ into the equation of the curve and equate it to the line equation.

When $m = 1$, the curve is $y = 2x^2 - 8x + 3$.
Equating the curve and the line $y = 1 - 4x$:
$2x^2 - 8x + 3 = 1 - 4x$
$2x^2 - 4x + 2 = 0$
$x^2 - 2x + 1 = 0$

2. Use the discriminant to show that the line is a tangent.

For the equation $2x^2 - 4x + 2 = 0$:
Discriminant $\Delta = b^2 - 4ac$
$\Delta = (-4)^2 - 4(2)(2)$
$\Delta = 16 - 16 = 0$
Since the discriminant is $0$, the line intersects the curve at exactly one point, hence it is a tangent to the curve.

3. Find the coordinates of the point of contact.

Solving $x^2 - 2x + 1 = 0$:
$(x - 1)^2 = 0$
$x = 1$
Substitute $x = 1$ into the line equation $y = 1 - 4x$:
$y = 1 - 4(1) = -3$
The coordinates of the point of contact are $(1, -3)$.

1. Set up the inequality for the curve to be above the line

For the curve to be above the line $y = 5$ for all real $x$, we require:
$(m+1)x^2 - 8x + 3m > 5$
$(m+1)x^2 - 8x + (3m - 5) > 0$

2. State the conditions for the quadratic expression to be always positive

For a quadratic $ax^2 + bx + c > 0$ for all $x$, we must have:

1. $a > 0 \implies m + 1 > 0 \implies m > -1$
2. Discriminant $b^2 - 4ac < 0$

3. Set up and simplify the discriminant inequality

$(-8)^2 - 4(m+1)(3m-5) < 0$
$64 - 4(3m^2 - 5m + 3m - 5) < 0$
$16 - (3m^2 - 2m - 5) < 0$
$16 - 3m^2 + 2m + 5 < 0$
$-3m^2 + 2m + 21 < 0$
$3m^2 - 2m - 21 > 0$

4. Solve the quadratic inequality in $m$

Factorising the quadratic:
$(3m + 7)(m - 3) > 0$
The critical values are $m = -\frac{7}{3}$ and $m = 3$.
Since the parabola opens upwards and we want the region $> 0$, the solution is:
$m < -\frac{7}{3} \quad \text{or} \quad m > 3$

5. Combine the conditions to find the final range of $m$

We must satisfy both $m > -1$ and $(m < -\frac{7}{3} \text{ or } m > 3)$.
Taking the intersection of these intervals, we get:
$m > 3$