Here is my proposition : a) Since $\forall x \in [0,1] : f(x) >0$ and $f$ is continuous then $u=\int_{0}^{1} f(t) dt >0$, let $n_{0}$ such taht $u > \frac{1}{n_{0}}$. Let $g(x)=\int_{0}^{x} f(t) dt$ , $g$ is an increasing and differentiable funciton, $g: [0,1] \to [0,u]$. hence for all $n\geq n_{0}$ we can find $x_{n}\in (0,1)$ such that $g(x_{n})=\frac{1}{n}$ b) By the Mean value theorem, we can find $t_{n} \in [0, x_{n}]$ such that $g(x_{n})-g(0)=x_{n} f(t_{n})$ and so $nx_{n}=\frac{1}{f(t_{n})}~~(*)$ , let $m=\inf_{x \in [0,1]} (f(x))$ we have $$ 0 \leq m x_{n} \leq \int_{0}^{x_{n}} f(t)dt =\frac{1}{n}$$ And so $t_{n} \to 0$ which give from $(*)$ $$\lim~~ n x_{n} =\frac{1}{f(0)}$$

Good problem