= Solution
Write $d=2n$ with $n\geq2$, and let $E$ be the <splitting field>. The root set of $X^{q^{2n}}+X^{q^2}+tX$ is an $n$-dimensional <vector space> over $\mathbb F_{q^2}$, by the argument of part (a) with $q$ replaced by $q^2$.
The constants $\mathbb F_{q^2}$ really occur inside $E$, rather than only in the ambient <algebraic closure>. Choose a nonzero root $\alpha$. For every $c\in\mathbb F_{q^2}$, $c\alpha$ is also a root, so both elements lie in $E$ and
$$
c=\frac{c\alpha}{\alpha}\in E.
$$
Let $K=\mathbb F_{q^2}(t)$. Since $K/\mathbb F_q(t)$ is Galois of degree two,
$$
H=\operatorname{Gal}(E/K)\triangleleft G,
\qquad\boxed{[G:H]=2.}
$$
This index statement does not require proving that $E$ has no still larger constant <field>.
As an extension of $K$, $E$ is the <splitting field> of the same <polynomial>. Applying part (b) with base constant size $q^2$ and dimension $n$ gives
$$
SL(n,q^2)\leq H\leq GL(n,q^2).
$$
The <special linear group> is the kernel of determinant in the <general linear group>, so every element of $H$ normalizes it. Therefore \b[$H$ is the required normal index-two subgroup normalizing $SL(d/2,q^2)$].
In fact the full group is semilinear over $\mathbb F_{q^2}$: an element outside $H$ acts on constants as $c\mapsto c^q$ and satisfies $g(cv)=c^qg(v)$. <Field> automorphisms preserve determinant one, so this also explains the <normalizer> behavior geometrically. This is the <quadratic constant extension of a q-squared linearized splitting field>, not the <polynomial> of part (b) with its $X^q$ term unchanged.
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