Here is how I answered the question. Could you please say what is not correct if it's incorrect?
Taking $ϵ = 1$ in the definition of the uniform convergence, we find that there exists $N ∈ N $such that $|f_{n}(x) − f(x)| < 1$ for all $x ∈ A $ if $p > q$. Choose some $n > N$. Then, since $f_n$ is bounded, there is a constant $Mn ≥ 0$ such that $|f_n(x)| ≤ Mn$ for all $x ∈ A$. It follows that $|f(x)| ≤ |f(x) − f_n(x)| + |f_n(x)| < 1 + Mn$ for all $x ∈ A$, meaning that $f$ is bounded on $A $(by $1 + Mn$). If $ 0 < x ≤ 1$, then $fn(x) = 0$ for all $ n ≥ 1/x$, so $ f_n(x) → 0$ as $n → ∞$; and if $x = 0$, then $f_n(x) = 0$ for all $n$, so $f_n(x) → 0 $ also. It follows that $f_n → 0$ pointwise on $[0, 1]$. This is the case even though $||f_n|| = n → ∞ $ as $n → ∞$. Thus, a pointwise convergent sequence of functions need not be bounded, even if it converges to zero. As the boundaries are not set within the exclusive set of range values that satisfies the condition of $p,q>0 $ thus the convergence of the range uniformly sets on the interval of the $[0.01, 1] $
Remark: above you have written that $f_n(x) = 0$ for all $n \geq 1/x$. But $f_n(x) = n^p x \exp(-n^q x) > 0$ for all $x > 0$. To prove the uniform convergence it is sufficient to use the following equivalence: $$ f_{n} \rightarrow f \; (\mbox{uniformly}) \iff \| f_n - f \|_{\infty} \rightarrow 0 \; \mbox{when } n \rightarrow + \infty. $$ In your case, we have $\|f_n\|_{\infty} = n^{p - q} e^{-1}$ which goes to zero when $n \rightarrow + \infty$. So, if $q > p$ we have uniform convergence. On the other hand, if $p > q$, then $n^{p - q} \rightarrow + \infty$ as $n \rightarrow + \infty$ (If $p = q$, $\|f_n\|_{\infty} = \frac{1}{e}$). Hence, in this case, using the above equivalence, we conclude that the convergence cannot be uniform.