Both parts of this are actually quite tricky.

Good start. So we now want to show that if 0<x<1 then $\displaystyle nx^n\to0$ as $\displaystyle n\to\infty$. As you can see, this is not obvious. One sneaky way to prove it is to notice that the series $\displaystyle \sum_{n=0}^\infty nx^n$ converges if |x|<1 (easily shown by using the ratio test), and therefore the n'th term of the series must tend to 0.

To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of $\displaystyle x^n(1-x)$ in the unit interval occurs when the derivative is 0. This occurs when $\displaystyle x=1-\tfrac 1{n+1}$, at which point the value of the function is $\displaystyle \bigl(\tfrac n{n+1}\bigr)^{n+1}$, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.

As I said, it's a tricky question.