The source code of this answer is available at my github gist.
I use a quite different method from yours.
For (1), we see that
\begin{equation}
\begin{aligned}
I=&\int_{|x|\leq 1/2}|H_t(x)|^2 d x\\
=&\int_{|x|\leq 1/2}H_t(x)\overline{H_t(x)}d x\\
=&\int_{|x|\leq 1/2}\sum_{n}e^{-4\pi^2n^2t}e^{2\pi i n x}\sum_{k}e^{-4\pi^2k^2t}e^{-2\pi i k x}\\
=& \sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int_{|x|\leq 1/2}e^{2\pi i (n-k)x}d x\\
=& \sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\mathbb{I}({n=k})\\
=&\sum_{n}e^{-8\pi^2n^2t}
\end{aligned}
\end{equation}
Note that
\begin{equation}
\begin{aligned}
I=&\sum_{n}e^{-8\pi^2n^2t}=1+2\sum_{n=1}^{\infty}e^{-8\pi^2n^2t}\\
\leq& 1+2 \sum_{n=1}^{\infty}\int_{n-1}^n e^{-8\pi^2 x^2 t}d x=1+2\int_{0}^{\infty}e^{-8\pi^2 x^2 t}d x\\
=&1+\int_{-\infty}^{+\infty}e^{-8\pi^2 x^2 t}d x
\end{aligned}
\end{equation}
by the same trick, we can show that
\begin{equation}
\begin{aligned}
I=&\sum_{n}e^{-8\pi^2n^2t}=-1+2\sum_{n=0}^{\infty}e^{-8\pi^2n^2t}\\
\geq & -1+2 \sum_{n=0}^{\infty}\int_{n}^{n+1} e^{-8\pi^2 x^2 t}d x=-1+2\int_{0}^{\infty}e^{-8\pi^2 x^2 t}d x\\
=&-1+\int_{-\infty}^{+\infty}e^{-8\pi^2 x^2 t}d x
\end{aligned}
\end{equation}
and by the Normal distribution $N(0,\sigma^2)$, we see that (the constant might be wrong, but it does not matter.)
\begin{equation}
\int_{-\infty}^{+\infty}e^{-8\pi^2 x^2 t}d x=\int\exp\left(-\frac{x^2}{2\frac{1}{16\pi^2t}}\right)d x=\sqrt{2\pi t }\cdot 4\pi=O(t^{-1/2}).
\end{equation}
Thus $I=O(t^{-1/2})$.
For (2),by the hint we have $c_1\sin^2(\pi x)\leq x^2\leq C_1 \sin^2(\pi x)$, for $|x|\leq 1/2$, i.e., $x^2=O(\sin^2(\pi x))$, we write $x^2=C \sin^2(\pi x)$.
\begin{equation}
\begin{aligned}
I_2=&\int_{|x|\leq 1/2}x^2|{H_t(x)}|^2d x=C\int\sin^2(\pi x)|H_t(x)|^2d x\\
=&C\int \frac{1}{2}(1-\cos(2\pi x))\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}e^{2\pi i (n-k)x} d x\\
=&\frac{C}{2}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n-k)x}\left(1-\frac{e^{2\pi i x}+e^{-2\pi i x}}{2}\right)d x\\
=&\frac{C}{2}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n-k)x}d x\\
&-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n+1-k)x}d x\\
&-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\int e^{2\pi i (n-1-k)x}d x\\
=&\frac{C}{2}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\mathbb{I}({n=k})\\
&-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\mathbb{I}({n=k-1})\\
&-\frac{C}{4}\sum_{n}\sum_{k}e^{-4\pi^2(n^2+k^2)t}\mathbb{I}({n=k+1})\\
=&\frac{C}{2}\sum_{n}e^{-8\pi^2n^2t}-\frac{C}{4}\sum_{n}\left[e^{-4\pi^2(2n^2+2n+1)t}+e^{-4\pi^2(2n^2-2n+1)t}\right]\\
=&\frac{C}{2}\sum_{n}e^{-8\pi^2n^2t}\left[1-\frac{e^{-4\pi^2(1+2n)t}+e^{-4\pi^2(1-2n)t}}{2}\right]\\
=&\frac{C}{2}\sum_{n}e^{-8\pi^2n^2t}\left[1-e^{-4\pi^2t}\frac{e^{-8\pi^2nt}+e^{8\pi^2nt}}{2}\right]\\
\end{aligned}
\end{equation}
Note that $2\leq e^x +e^{-x}\leq 2 e^{|x|}$
when $0\leq x \leq \delta$, $e^x \sim 1+x$, one can see
\begin{equation}
1-e^{-4\pi^2t}\frac{e^{-8\pi^2nt}+e^{8\pi^2nt}}{2}\leq 1-e^{-4\pi^2t}\leq c 4\pi^2t .
\end{equation}
and
\begin{equation}
1-e^{-4\pi^2t}\frac{e^{-8\pi^2nt}+e^{8\pi^2nt}}{2}\geq 1-e^{-4\pi^2t}e^{8\pi^2|n|t}\geq c 4\pi^2(2|n|-1)t .
\end{equation}
Finally, we have
\begin{equation}
\begin{aligned}
I_2\leq K t \sum_{n}e^{-8\pi^2n^2t}=t O(t^{-1/2})=O(t^{1/2}),
\end{aligned}
\end{equation}
and
\begin{equation}
I_2\geq Kt \sum_{n}e^{-8\pi^2n^2t}(2|n|-1)=2Kt \sum_{n}|n|e^{-8\pi^2n^2t} -O(t^{1/2}) =O(t^{1/2}).
\end{equation}
Note that
\begin{equation}
\sum_{n}|n|e^{-8\pi^2n^2t} \sim\int |x|e^{-8\pi^2x^2t }d x=O(t^{-1/2}),
\end{equation}
which is $E(|X|)$ for $X\sim N(0,\sigma^2)$, and $E(|X|)=\sigma\sqrt{\frac{2}{\pi}}$. see this question for more details.
By far we have proved that $I_2=O(t^{1/2})$.
