= Solution
A centered random variable is <sub-Gaussian random variable>[sub-Gaussian] with parameter $\sigma^2$ when
$$
\mathbb Ee^{\lambda X}\leq e^{\sigma^2\lambda^2/2}
\qquad(\lambda\in\mathbb R).
$$
For $\lambda>0$, the <Chernoff bound> gives
$$
\mathbb P(X\geq x)\leq
\exp\left(-\lambda x+\frac{\sigma^2\lambda^2}{2}\right).
$$
Minimizing at $\lambda=x/\sigma^2$ yields $e^{-x^2/(2\sigma^2)}$. Applying the same argument to $-X$ proves the left-tail bound.
The moment-generating-function inequality and its version at $-\lambda$ imply
$$
\mathbb E\cosh(\lambda X)\leq e^{\sigma^2\lambda^2/2}.
$$
Comparing the second-order terms as $\lambda\to0$ gives $\mathbb EX^2\leq\sigma^2$. Since $\mathbb EX=0$,
$$
\operatorname{Var}(X)\leq\sigma^2.
$$
A centered $Z$ is <Sub-Gamma random variable in the right tail> with variance factor $v$ and scale factor $c$ when
$$
\log\mathbb Ee^{\lambda Z}
\leq\frac{v\lambda^2}{2(1-c\lambda)}
\qquad(0<\lambda<c^{-1}).
$$
The corresponding <Bernstein's inequality> is
$$
\mathbb P(Z\geq x)
\leq\exp\left(-\frac{x^2}{2(v+cx)}\right).
$$
A standard squared-sub-Gaussian lemma, obtained by integrating the sub-Gaussian tail or expanding exponential moments, states that
$$
X^2-\mathbb EX^2\in\Gamma_+(16\sigma^4,2\sigma^2),
\qquad
\mathbb EX^2-X^2\in\Gamma_+(16\sigma^4,\sigma^2).
$$
Scaling a sub-Gamma variable by $a\geq0$ multiplies its variance factor by $a^2$ and its scale by $a$; independent sums add variance factors and take the largest scale. Decompose
$$
a_i(X_i^2-\mathbb EX_i^2)
=a_i^+(X_i^2-\mathbb EX_i^2)
+a_i^-(\mathbb EX_i^2-X_i^2).
$$
Their sum is therefore sub-Gamma on the right with variance factor
$$
\sigma^4v
=16\sigma^4\{\lVert a^+\rVert_2^2+\lVert a^-\rVert_2^2\}
$$
and scale factor
$$
\sigma^2c
=\sigma^2\max\{2\lVert a^+\rVert_\infty,\lVert a^-\rVert_\infty\}.
$$
Apply Bernstein to the sum at threshold $nx$ to obtain
$$
\boxed{
\mathbb P\left\{\frac1n\sum_{i=1}^na_i(X_i^2-\mathbb EX_i^2)\geq x\right\}
\leq
\exp\left[-\frac{n^2x^2}{2\{\sigma^4v+cn\sigma^2x\}}\right]}.
$$
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