If there is no smallest root,
then, for any $c > 0$,
there is a root $r$
such that
$0 < r < c$.
Use this to prove that
$f$ is not continuous at zero.
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 $g.$
Best Answer
No, there is no such function. Suppose that $F(x)+F(x^2)=x$ and $F$ is continuous on $[0,1]$. We use the function $f(x), \ g(x)$ in my answer to this post where it is proved that $f(x)-g(x)$ is not a constant function. The functions $f(x), \ g(x)$ originate from Hardy's 'Divergent Series'. To recall the definition, $$ f(x)=\sum_{n=0}^{\infty} (-1)^n x^{2^n}, $$ $$ g(x)=\sum_{n=0}^{\infty} \frac{(\log x)^n}{(2^n+1)n!}. $$ It is easy to check that $f(x)+f(x^2)=x$ and $g(x)+g(x^2)=x$.
Now, since $F(x)$ on $[0,1]$ is continuous and $F(x)+F(x^2)=x$, we have $\Phi(x)=F(x)-g(x)$ satisfying $$ \Phi(x)=-\Phi(x^2)=\Phi(x^4). $$ Note that $g(x)$ is continuous on $(0,1]$. Thus, $\lim_{x\rightarrow 1-} \Phi(x)$ exists. Then such $\Phi$ satisfying $\Phi(x)=\Phi(x^4)$ must be a constant function. Let $\Phi(x)=c$. We have $$ F(x)-g(x)=c \ \ \textrm{ if } x\in (0,1]. \ \ (1) $$ Similarly, let $\Psi(x)=f(x)-F(x)$. Then $\Psi(x)=\Psi(x^4)$. As $f$ is a power series at $0$, we have $\lim_{x\rightarrow 0+} f(x) = 0$. Then we must have $\lim_{x\rightarrow 0+} \Psi(x)=-F(0)$. Such $\Psi$ satisfying $\Psi(x)=\Psi(x^4)$ must be a constant function. Thus, $$ f(x)-F(x)=-F(0) \ \ \textrm{ if } x\in [0,1). \ \ (2) $$ Summing up (1) and (2), we have $$ f(x)-g(x)=c-F(0) \ \ \textrm{ if } x\in (0,1). $$ Therefore, we have a contradiction since $f(x)-g(x)$ cannot be constant on $(0,1)$ as proved in my MO answer.
Remark The phenomenon we are observing here is that $f$ is oscillatory around $1$, and $g$ is oscillatory around $0$.