Is $\ n\cdot (1-x)\cdot e^{-n\cdot(1-x)}$ uniformly and pointwise convergent in [0, a]? 
Is $\ n\cdot (1-x)\cdot e^{-n\cdot(1-x)}$ uniformly and pointwise
  convergent in [0, a]?

$\ f_n:[0, a]\to \mathbb{R}, 0<a\le 1 $
I started by calculating the limit of $\ f_n(0)$:
$\ \lim_{n\to \infty} f_n(0)= n\cdot (1-0)\cdot e^{-n\cdot(1-0)}=n\cdot e^{-n}$.
How do I continue now?..
 A: You have $f_n(1)=0$. When $0\leq x<1$, $$\lim_{n\to\infty}f_n(x)=0,$$ since the exponential with negative exponent wins over the linear factor. 
Assume first that $a<1$. For $x\leq a$ we have $1-x\geq 1-a$, so 
$$
0\leq n(1-x)e^{-n(1-x)}\leq n(1-x) e^{-n(1-a)}\leq n e^{-n(1-a)}
$$
and the convergence is uniform. 
When $a=1$, though, the convergence is not uniform. If we let 
$$
x_n=1-\tfrac1n,
$$
then for all $n$
$$
f_n(x_n)=n(1-x_n)\,e^{-n(1-x_n)}=e^{-1}.
$$
So there is no possible $N$ such that $|f_n(x)|<e^{-1}$ for all $x\in[0,1]$ and $n>N$. 
A: For any $x\in[0,1]$ we can easily see that $$\lim_{n\to\infty} \frac{n(1-x)}{e^{n(1-x)}}=0$$ and thus $f_n\to 0$ pointwise on $[0,1]$. Notice that $f_n'(x)=-n(nx-n+1)e^{-n(1-x)}$ so the critical points are the roots of $f_n'$, i.e. $x_n=\frac{n-1}{n}$. The height at this critical point of $f_n$ is $f_n\left(x_n\right)=\frac{1}{e}\approx. 0.368$, and one can verify that this is the global maximum of $f_n$ on $[0,1]$. For each $0<a<1$, we have $x_n\to 1>a$ so that for large $n$ we have $x_n>a$. Each $f_n$ is increasing on $[0,x_n]$, so for large $n$ we have $f_n$ increasing on $[0,a]$. Thus, for large $n$, the maximum of $f_n$ on $[0,a]$ is at $x=a$, and $f_n(a)\to 0$. Thus, for each $0<a<1$, we have $f_n\to 0$ uniformly on $[0,a]$, however by looking at the maxima $x_n$ we easily see that $f_n$ does not converge uniformly on $[0,1]$. 
