Solution to 1
Let $H=\left\{ \begin{bmatrix} 1 & 0 & 0 & ...\\0 \\... & & M\\0 \end{bmatrix} : M \in GL_{n-1}(F) \right\}$
Take $a,b \in H$, say $M=A$ for $a$ and $M=B$ for $b$. Now $ab^{-1}= \begin{bmatrix} 1 & 0 & 0 & ...\\0 \\... & & AB^{-1}\\0 \end{bmatrix} \in H$
So by abbreviated subgroup test, $H \le G$. Now let $\phi:H \rightarrow GL_{n-1}(F)$ where phi takes $a\mapsto A$. This is clearly a bijection since the extra entries are just 0 and 1 and are always the same.
$\phi(ab) = \phi \bigg( \begin{bmatrix} 1 & 0 & 0 & ...\\0 \\... & & AB\\0 \end{bmatrix} \bigg) = \begin{bmatrix} AB \end{bmatrix} =\begin{bmatrix} A \end{bmatrix}\begin{bmatrix} B \end{bmatrix} = \phi(a)\phi(b)$. So yes $\phi$ is an isomorphism.
So $H \cong GL_{n-1}(F)$. $\Box$
Solution to 2
Let $N=\left\{ \begin{bmatrix} 1 & x_1 & x_2 & ...\\0 \\... & & I_{nā1}\\0 \end{bmatrix} : x_i \in F \forall i \right\}$
Note: $X = \begin{bmatrix} x_1 &x_2 &... \end{bmatrix} \Rightarrow X \in F^{nā1}$
I figured it out. Showing subgroup straightforward. Normalcy is also not bad.
Take $\psi:N\rightarrow F^{n-1}$ where $n\in N\mapsto X$. Bijection. $\psi(ab)=[A+B]=[A]+[B]=\psi(a)+\psi(b)$. $\Box$
Solution to 3
First of all, $H \bigcap N = I_n$ is pretty obvious by the construction.
Next, $NH = \begin{bmatrix} 1 & x_1 & x_2 & ...\\0 \\... & & M \\0 \end{bmatrix}$ is the entirety of $G$.
Clearly $NH \subset G$ and $NH \supset G$ $\Rightarrow G=NH$
Then, since $H \bigcap N = I_n$ and $G = NH$, we have 3 done by my professor's construction of semi-direct product.