1. Find set $A$

The quadratic is $x^2-(m+1)x+(m+4)=0.$

For both roots to be real, its discriminant must satisfy $\Delta \ge 0.$

Here, $\Delta=(m+1)^2-4(m+4).$

Now simplify: $\Delta=m^2+2m+1-4m-16=m^2-2m-15.$

Factorize: $m^2-2m-15=(m-5)(m+3).$

So the condition is $(m-5)(m+3)\ge 0.$

This gives $m\le -3 \quad \text{or} \quad m\ge 5.$

Hence, $A=(-\infty,-3]\cup[5,\infty).$

Given $B=[-3,5).$

2. Check each option

Option A: $A\cap B=\{-3\}$

Compute the intersection: $A\cap B=\left[(-\infty,-3]\cup[5,\infty)\right]\cap[-3,5).$

• $(-\infty,-3]\cap[-3,5)=\{-3\}$
• $[5,\infty)\cap[-3,5)=\varnothing$

Therefore, $A\cap B=\{-3\}.$

So Option A is true.

Option B: $B-A=(-3,5)$

Now remove from $B=[-3,5)$ all elements that are in $A$.

Since only $-3$ from $B$ belongs to $A$, $B-A=[-3,5)\setminus\{-3\}=(-3,5).$

So Option B is true.

Option C: $A\cup B=\mathbb{R}$

We have $A=(-\infty,-3]\cup[5,\infty), \qquad B=[-3,5).$

Their union is $(-\infty,-3]\cup[-3,5)\cup[5,\infty)=\mathbb{R}.$

So Option C is true.

Option D: $A-B=(-\infty,-3)\cup(5,\infty)$

Compute: $A-B=A\setminus B.$

Now,

• from $(-\infty,-3]$, the point $-3$ is in $B$, so it is removed, giving $(-\infty,-3)$;
• from $[5,\infty)$, since $5\notin B$ (because $B=[-3,5)$), the entire $[5,\infty)$ remains.

Thus, $A-B=(-\infty,-3)\cup[5,\infty).$

But the option states $(-\infty,-3)\cup(5,\infty),$ which incorrectly excludes $5$.

So Option D is not true.

3. Final conclusion

The statement which is not true is: $\boxed{\text{D}}$