Question 1: $X_n$'s are non-negative, uniformly integrable. Then $E\left[\dfrac{\max_{1\leq k \leq n} X_k}{n}\right]\rightarrow 0$.
Question 2: If u.i. is dropped then the above may fail.
My thought: If I can show $\dfrac{\max_{1\leq k \leq n} X_k}{n}$ is u.i. as well as converges in probability to $0$ then I am done. Probably I am not thinking in the right direction. Any kind of help/hint is what I am looking for and will appreciate. Thank you.
Q1: You can prove it directly, without the Vitali convergence theorem.
A hint to get you started:
Let's let $M_n = \max_{1 \le k \le n} X_k$. Now fix $\epsilon > 0$ and write $$E\left[\frac{M_n}{n}\right] \le E\left[\frac{M_n}{n}; M_n \le \epsilon n\right] + \sum_{k=1}^n E\left[\frac{M_n}{n}; M_n > \epsilon n, M_n = X_k\right].$$ (The inequality is to handle the possibility of a "tie", when $M_n = X_k$ for more than one $k$, so that the events $\{M_n = X_k\}, k=1,\dots,n$ are not disjoint.)
Q2: As in my comment, if you choose the $X_k$ to have disjoint support, then $M_n = X_1 + \dots + X_n$. Now choose them such that $E[X_k] = 1$ for every $k$. (You can construct such a sequence on the probability space $[0,1]$ equipped with Lebesgue measure...)