1. We need to study the number of real roots of $f(x)=x^5-5x+a.$ Let $g(x)=x^5-5x,$ so that solving $f(x)=0$ is equivalent to solving $g(x)=-a.$ Thus, the number of real roots depends on how many times the horizontal line $y=-a$ cuts the curve $y=g(x)$.

2. Find the critical points of $g$. $g'(x)=5x^4-5=5(x^4-1)=5(x^2-1)(x^2+1).$ Hence the real critical points are $x=\pm 1.$

3. Determine increasing/decreasing behavior.

• If $|x|>1$, then $x^4>1$, so $g'(x)>0$.
• If $|x|<1$, then $x^4<1$, so $g'(x)<0$.

Therefore:

• $g$ is increasing on $(-\infty,-1)$,
• decreasing on $(-1,1)$,
• increasing on $(1,\infty)$.

So:

• $x=-1$ is a local maximum,
• $x=1$ is a local minimum.

4. Compute the extreme values. $g(-1)=(-1)^5-5(-1)=-1+5=4,$ $g(1)=1-5=-4.$ So the graph has:

• local maximum value $4$ at $x=-1$,
• local minimum value $-4$ at $x=1$.

5. Now solve graphically using $g(x)=-a$. We compare the level $-a$ with the interval $[-4,4]$.

Case 1: $a>4$

Then $-a<-4.$ The horizontal line $y=-a$ lies below the minimum value $-4$ of the middle turning region. So it cuts the curve only once. Hence $f(x)=0$ has only one real root.

Therefore:

• A is false,
• B is true.

Case 2: $a<-4$

Then $-a>4.$ The horizontal line $y=-a$ lies above the local maximum value $4$. So again it cuts the curve only once. Hence $f(x)=0$ has only one real root.

Therefore:

• C is false.

Case 3: $-4<a<4$

Then $-4<-a<4.$ The horizontal line $y=-a$ lies strictly between the local minimum and local maximum values. So it cuts the curve in three distinct points. Hence $f(x)=0$ has three real roots.

Therefore:

• D is true.

6. (Boundary check) If $a=4$, then $-a=-4$, giving a repeated root at $x=1$. If $a=-4$, then $-a=4$, giving a repeated root at $x=-1$. These are not included in the strict inequalities, so no issue.

7. Final evaluation of options:

• A: False
• B: True
• C: False
• D: True

Thus the correct options are $\boxed{B, D}.$