Let us define $g(x)=f(x)-(2x+3).$ Then the given line $y=2x+3$ intersects the graph of $f$ exactly where $g(x)=0$.

We are given: $f(0)=3,\, f(1)=5.$ So, $g(0)=f(0)-3=0,$ and $g(1)=f(1)-(2+3)=0.$

Also, the line intersects the graph of $f$ at only two distinct points in $(0,1)$. Hence, in the open interval $(0,1)$, the equation $g(x)=0$ has exactly two distinct roots. Let them be $a$ and $b$ with $0<a<b<1.$

Thus, $g(x)$ has the four distinct zeros: $0,\ a,\ b,\ 1.$

Now we apply Rolle's theorem repeatedly.

1. Zeros of $g'$

Since $g$ is differentiable on each closed subinterval and continuous everywhere, by Rolle's theorem:

• on $[0,a]$, there exists $c_1\in(0,a)$ such that $g'(c_1)=0;$
• on $[a,b]$, there exists $c_2\in(a,b)$ such that $g'(c_2)=0;$
• on $[b,1]$, there exists $c_3\in(b,1)$ such that $g'(c_3)=0.$

These are three distinct points because they lie in disjoint intervals.

2. Zeros of $g''$

Now $g'(c_1)=g'(c_2)=g'(c_3)=0$. Again by Rolle's theorem:

• on $[c_1,c_2]$, there exists $d_1\in(c_1,c_2)$ such that $g''(d_1)=0;$
• on $[c_2,c_3]$, there exists $d_2\in(c_2,c_3)$ such that $g''(d_2)=0.$

So there are at least two distinct points in $(0,1)$ where $g''(x)=0.$

But $g''(x)=f''(x)-0=f''(x).$ Hence, there are at least two distinct points in $(0,1)$ such that $f''(x)=0.$

3. Least number

The above argument guarantees at least 2 such points. This bound is also the least possible.

Therefore, the least number of points is $\boxed{2}.$

4. Comparison with stored answer

Stored correct answer: $2$.

Our derived answer matches the stored answer.