Q1: The answer is no. If $g''(a)>0$ for some $a\in [0,1],$ then $g''>0$ in a neighborhood of $a$ by the continuity of $g''.$ Hence $g$ is strictly convex in that neighborhood. Similarly, if $g''(a)<0$ for some $a\in [0,1],$ then $g$ is strictly concave in that neighborhood. We're left with the case $f''\equiv 0.$ But this implies $f(x) = ax +b$ on $[0,1],$ hence $f$ is both convex and concave everywhere on $[0,1].$
Added later, in answer to the comment: It's actually possible for a $C^2$ function to have uncountably many inflection points. Suppose $K\subset [0,1]$ is uncountable, compact, and has no interior (the Cantor set is an example). Define
$$f(x)=\begin{cases}d(x,K)\sin (1/d(x,K)),&x\notin K\\ 0,& x\in K\end{cases}$$ Then $f$ is continuous, and $f$ takes on positive and negative values on any interval containing a point of $K.$ Define
$$g(x)=\int_0^x\int_0^t f(s)\,ds\,dt.$$
Then $g\in C^2[0,1]$ and $g''=f.$ It follows that every point of $K$ is an inflection point of $g.$
The answer is positive.
Since the problem is invariant under the change $\lambda_n \iff 1-\lambda_n$, we may assume that $c=(1-\lambda_n)a_n+\lambda_n b_n$.
First, we note that $c= a_n +\lambda_n(b_n-a_n)\ge a_n +\lambda_n \epsilon $.
Now, set $\tilde a_n=c-\lambda_n \epsilon \ge a_n \ge 0$
, and $\tilde b_n=c+\epsilon(1-\lambda_n )$.
Then $$c=(1-\lambda_n)\tilde a_n+\lambda_n \tilde b_n=(1-\lambda_n)a_n+\lambda_n b_n$$, and since $\tilde a_n \ge a_n$ it follows that $\tilde b_n \le b_n$.
The convexity of $F$ now implies that
$$
(1-\lambda_n)F(\tilde a_n)+\lambda_n F(\tilde b_n) \le(1-\lambda_n)F(a_n)+\lambda_n F(b_n),
$$
so $\tilde D_n \le D_n$ where
$$\tilde D_n:=\lambda_nF(\tilde a_n)+(1-\lambda_n)F(\tilde b_n)-F\big((1-\lambda_n) \tilde a_n +\lambda_n\tilde b_n\big).$$
So we replaced $(a_n,b_n)$ with bounded sequences $(\tilde a_n,\tilde b_n)$ satisfying $\tilde b_n-\tilde a_n =\epsilon$, (while keeping the constants $\lambda_n$ the same). This reduces the problem to the case where $b_n$ is bounded, so we are done.
In fact, we can get even a quantitative estimate on how fast the $\lambda_n$ must converge:
Now, $[\tilde a_n,\tilde b_n] \subseteq [\min(c-\epsilon,0),c+\epsilon)$, and $\tilde b_n-\tilde a_n =\epsilon$.
Thus,
$$
D_n \ge \tilde D_n \ge 1/2 (\min_{x \in [\min(c-\epsilon,0),c+\epsilon]} F''(x)) \lambda_n(1-\lambda_n)(\tilde b_n-\tilde a_n)^2=$$ $$1/2 (\min_{x \in [\min(c-\epsilon,0),c+\epsilon]} F''(x)) \epsilon^2.
$$
Best Answer
The idea is to "go one step" further in the chain of derivatives, and to understand what properties do we need from $f'$, in order to get an example of a non-convex function with a strict minimum.
Let $g:[0,1) \to \mathbb R$ be a smooth function satisfying
Such functions exist. Given such a function $g$, define $h:(-1,1) \to \mathbb R$ by $$ h(x)= \begin{cases} g(x), & \text{ if }\, x \ge 0\\ -g(-x), & \text{ if }\, x \le 0 \end{cases}$$
Property $(3)$ of $g$ implies that $h$ is smooth. Note that $h>0$ on $(0,1)$ and $h<0$ on $(-1,0)$.
Finally, define $f(x)=\int_0^x h(t)dt$. Then $$ f''(x)= \begin{cases} g'(x), & \text{ if }\, x \ge 0\\ g'(-x), & \text{ if }\, x \le 0 \end{cases}$$ obtains negative values arbitrarily close to $0$, by property $(2)$ of $g$.