No. Think of a sequence of $f_n$ such that $\| f_n\|_2=1$ while $\|g f\|_2\to 0$. And you might as well assume $g(x)=\sqrt x$. Which is equivalent to asking the same question with $f_n\geq0$, $\| f_n\|_1 =1$, $\| xf_n\|_1\to0$ (just take $g_n=\sqrt{f_n}$ for the $L^2$ case).

Let us try $f_n=\begin{cases} n & \textrm{when } x<\frac1n\\ 0 &\textrm{otherwise}\end{cases}.$ We compute $\int_0^1 x f_n \textrm{d} x = \frac{1}{2n}\to0$, and that's that.

No. Think of a sequence of $f_n$ such that $\| f_n\|_2=1$ while $\|g f\|_2\to 0$. And you might as well assume $g(x)= x^{\alpha}$.

It is equivalent to asking the same question with $f_n\geq0$, $\| f_n\|_1 =1$, $\| x^{2\alpha}f_n\|_1\to0$ (just take $g_n=\sqrt{f_n}$ for the $L^2$ case).

Let us try $f_n=\begin{cases} n & \textrm{when } x<\frac1n\\ 0 &\textrm{otherwise}\end{cases}.$ We check that it works: $$\int f_n \textrm{d} x =1,\quad \int_0^1 x^{2\alpha} f_n \textrm{d} x = \frac{1}{(2\alpha+1)n^{2\alpha}}\to0.$$