$\{(12)(34),(13)(24)\}$ is not a subgroup of $A_4$ because it does not contain identity element.
I think $N_1$ should be $$N_1=\{1,(12)(34),(13)(24),(14)(23)\}$$ Note that $N_1$ is a normal subgroup of $A_4$.
Now take $N_2=\{1,(12)(34)\}$ or $\{1,(13)(24)\}$ or $\{1,(14)(23)\}$
These are normal in $N_1$ because they are of index $2$ in $N_1$.
Symplectic group
I won't yet address identifying all maximal unipotent subgroups, but will just describe the standard ones.
The maximal unipotent subgroups of the symplectic group in four dimensions over the field $K$ have a nice description in terms of certain nice elements.
$\newcommand{\m}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}$
Let $$G=\left\{ g \in \operatorname{GL}(4,K) : gxg^T = x \right\}, \quad x=\m{0&0&0&1\\ 0&0&1&0\\ 0&-1&0&0\\ -1&0&0&0}$$
be a symplectic group. It has two standard maximal unipotent subgroups, the upper and lower, which differ only by being transposes of each other. The lower on is: $$P=\left\{ \m{ 1 & 0 & 0 & 0 \\ x & 1 & 0 & 0 \\ y & w & 1 & 0 \\ z & y-xw & -x & 1 } : x,y,z,w \in K \right\}$$
Define $$
x_1(t) = \m{1&0&0&0\\t&1&0&0\\0&0&1&0\\0&0&-t&1}, \quad
x_2(t) = \m{1&0&0&0\\0&1&0&0\\0&t&1&0\\0&0&0&1}, \quad
x_3(t) = \m{1&0&0&0\\0&1&0&0\\t&0&1&0\\0&t&0&1}, \quad
x_4(t) = \m{1&0&0&0\\0&1&0&0\\0&0&1&0\\t&0&0&1}, \quad
$$
Note that $$\begin{array}{lrl}
(1) & x_i(s)x_i(t) &= x_i(s+t) \\
(2) & x_2(t)x_1(s) &= x_1(s) x_2(t) x_3(st) x_4(s^2t) \\
(3) & x_3(t)x_1(s) &= x_1(s) x_3(t) x_4(2st) \\
(4) & x_4(t)x_1(s) &= x_1(s) x_4(t) \\
(5) & x_3(t)x_2(s) &= x_2(s) x_3(t) \\
(6) & x_4(t)x_2(s) &= x_2(s) x_3(t) \\
(7) & x_4(t)x_3(s) &= x_3(s) x_4(t) \\
\end{array}$$
If $K$ has characteristic not $2$, then $P$ has rank $2k$, generated by $x_1(s)$, and $x_2(t)$ where $s,t$ range over a generating set of the additive group of $K$. $[P,P]$ is isomorphic to the additive group of a two dimensional vector space $K^2$ over $K$, and is generated as a group by $x_3(s)$ and $x_4(t)$ with $s,t$ from a generating set of the additive group of $K$. $[P,P,P]$ is isomorphic to the additive group of $K$, and is generated as a group by $x_4(t)$ where $t$ ranges over a generating set of the additive group of $K$.
When $K$ has characteristic 2, things make less sense to me. The nilpotency class is 2. When $|K|>2$, $P/[P,P] \cong [P,P] \cong (K^+)^2$, but when $|K|=2$, $P\cong C_2 \times D_8$ behaves differently.
The normalizer of $P$ is the semidirect product of $P$ and a maximally split maximal torus $$H=\left\{ \m{ s & 0 & 0 & 0 \\ 0 & t & 0 & 0 \\ 0 & 0 & t^{-1} & 0 \\ 0 & 0 & 0 & s^{-1} } : s,t \in K^\times \right\}$$
In particular, the collection of maximal unipotent subgroups is in bijection with $(|K|^2+1)(|K|+1)^2$.
Best Answer
(a) $A_4$ is a group of order 12. Now, the subgroups of $A_4$ must have an order that divides 12. So, nontrivial subgroups have potential orders: 6,4,3, and 2. Since $K$ is the Unique 2-Sylow subgroup, it must be normal. The quotient group has order 12/4 = 3 and therefore is cyclic(even stronger than abelian).
(b) Now, inside of $K$ we know a cyclic subgroup of order 2 must exist. This can be seen from Cauchy's Theorem which states for any prime, p, that divides the order of a group, there exists an element of order p. We know that the subgroup generated by the element of order 2 must be normal inside $K$ as its index is 2.