It's not true.
Consider the orthonormal basis $\{e_n : n \in \mathbb{Z}\}$ for $L^2[0, 2\pi]$, where $e_n(t) = \frac1{\sqrt{2\pi}}e^{int}$.
Then for any $g \in L^2[0,2\pi]$ we have $\langle e_n, g\rangle \xrightarrow{n\to\infty} 0$ by the Riemann-Lebesgue lemma, but $\|e_n\|_{L^2} = 1, \forall n\in\mathbb{N}$.
For $h_n$ to be well defined you need a unique representation $x = a/b$ for $x \in \mathbb{Q}$. Presumably you mean that $a/b$ is in lowest terms where $(a,b) = 1$, that is $a$ and $b$ are coprime.
In this case, we have for a bounded interval $[\alpha, \beta]$ that $f_n(x)$ converges uniformly to $f(x) = x$ and $g_n(x)$ converges uniformly to
$$g(x) = \begin{cases}0, &x = 0 \text{ or }x \in \mathbb{I}& \\ b,& x \in \mathbb{Q} \text{ with } x = a/b, \,(a,b) =1 \end{cases}$$
The convergence $g_n \to g$ is uniform since $|g_n(x) - g(x) | = 1/n$.
Consider the convergence $f_ng_n \to fg$ on the bounded interval $[0,1]$.
We have
$$\tag{*}|f_n(x)g_n(x) - f(x)g(x)| = \begin{cases}x(1/n + 1/n^2), & x= 0 \text{ or }x \in \mathbb{I}\cap [\alpha,\beta]& \\ x(1/n + 1/n^2) + (xb)/n,& x \in \mathbb{Q}\cap [\alpha,\beta] \text{ with } x = a/b, \,(a,b) =1 \end{cases}$$
where $(xb)/n = a/n$.
For any $m \in \mathbb{N}$ the rational $a/b = m/(m+1)$ is in $[0,1]$. Thus $a$ is unbounded and the supremum of (*) is unbounded and does not converge to $0$ as $n \to \infty$. Therefore, convergence is not uniform. We can make a similar argument for other bounded intervals.
Best Answer
We use the following claim:
Indeed, we have $$\left|\int_a^bf_n(t)dt-\int_a^bf(t)dt\right|\leq (b-a)\sup_{a\leq x\leq b}|f_n(x)-f(x)|,$$ which converges to $0$ thanks to the uniform convergence on $[a,b]$.
In our case, we have $$\int_0^{\frac{\pi}2}f_n(x)dx=n\left[-\frac{(\cos x)^{n+1}}{n+1}\right]_0^{\frac \pi 2}=\frac n{n+1}\to 1,$$ whereas $f_n$ converges pointwise to $0$. This shows that the convergence cannot be uniform.