I suppose your confusion is due to the double representation of the semidirect product: internal vs external.
The semidirect product's theorem states that if you have a group $G$ having two subgroups $H,K < G$ such that $H$ is normal in $G$, $H \cap K = \{1_G\}$ and $G=HK$ then there's an isomorphism $G \cong H \rtimes_\psi K$, for a certain $\psi \colon K \to \text{Aut}(H)$.
By the theorem we can represent every element of $G$ as a pair $(h,k) \in H \times K$ (which is the support of the group $H \rtimes K$).
Consider the two subgroups $\bar H = \{(h,1_K) | h \in H\} \leq H \rtimes K$ and $\bar K = \{(1_H,k)|k \in K\} \leq H \rtimes K$, these subgroups correspond, via the isomorphism, to the subgroups $H$ and $K$ of $G$.
In $H \rtimes K$ we have that for every $h \in H$ and $k \in K$
$$(1_H,k) * (h,1_K) *(1_H,k)^{-1} = (\psi_k(h),1_K)$$
if we identify every $h \in H$ with its corresponding element $(h,1_K)$ and every $k \in K$ with $(1_H,k)$ then this equality become (internally in $G$)
$$k*h*k^{-1}=\psi_k(h)$$
The $\psi$ which determine the operation in the semidirect product is exactly the homomorphism sending every $k \in K$ in the (restriction to $H$ of the) automorphism $\psi_k \colon H \to H$ which send $h \in H$ in $khk^{-1}$ (this is clearly well defined because $H$ is normal in $G$.
Hope this help.
My answer only concerns part (c).
To define a semidirect product $Z_{15} \rtimes T$, for $T = Z_4, Z_2 \times Z_2$, you must pick a morphism $T \to \operatorname{Aut} Z_{15}$, which will represent the conjugation action of $T$ on $Z_{15}$.
I think the key here is to recognize that this morphism is characterized by the group structure of $G$, and is independent of the copy of $T$ that you've selected within $G$. The reason is that conjugation defines a morphism $G \to \operatorname{Aut} Z_{15}$. Since $Z_{15}$ is Abelian, this map factors through $G/Z_{15}$, so the conjugation action of $T$ is $T \cong G/Z_{15} \to \operatorname{Aut} Z_{15}$.
Therefore the problem of classifying the isomorphism types of the semidirect products is the same as that of classifying the "types" of morphisms $\varphi \colon T \to \operatorname{Aut} Z_{15}$, where two morphisms $\varphi_1$ and $\varphi_2$ are considered equivalent if there exists an automorphism $\alpha$ of $T$ such that $\varphi_2 = \varphi_1 \circ \alpha$.
In practice, this means enumerating the subgroups $H$ of $\operatorname{Aut} Z_{15} \cong Z_2 \times Z_4$ which could possibly be a homomorphic image of $T$, and then determining the "abstract" ways that $H$ can be obtained as a quotient of $T$.
In the present situation, there is only ever one abstract way to obtain $\{1\}$, $Z_2$, $Z_4$ (or $Z_2 \times Z_2$, depending on $T$) as a quotient of $T$, so the problem really amounts to enumerating the subgroups of order $1$, $2$ and $4$ of $\operatorname{Aut} Z_{15}$. There are five subgroups that can be a quotient of $Z_2 \times Z_2$, and six that can be a quotient of $Z_4$.
Best Answer
You could do it even easier. $2012$ is an even number so there is also the dihedral group $D_{1006}$.