Addressing your first question:
Denote the space of Schwartz functions, $S(\mathbb{R})$, and consider the dual space, $S(\mathbb{R})^*$. If $T$ is in $S(\mathbb{R})^*$, then $T: S(\mathbb{R}) \mapsto \mathbb{R}$ is a linear functional. If $\phi \in S(\mathbb{R})$, you often you will see $T$ acting on $\phi$ written as $
\langle T, \phi \rangle$.
Define $\delta$ to be such that:
$$
\langle \delta, \phi \rangle = \phi(0)
$$
Clearly, $\delta \in S(\mathbb{R})^*$. Also, note that if $f$ is a locally integrable function, you can define $T_f$ such that:
$$
\langle T_f, \phi \rangle = \int_{\mathbb{R}} f(x) \phi(x)dx
$$
and $T_f \in S(\mathbb{R})^*$.
Now then, the Fourier transform on $S(\mathbb{R})^*$ is defined as
$$
\langle FT, \phi \rangle = \langle T, F\phi \rangle
$$
Why? Consider when $f: \mathbb{R} \mapsto \mathbb{R}$ and $Ff$ "makes sense". Then:
\begin{align*}
\langle FT_f, \phi \rangle &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} f(y) \exp(-ixy)dy\right]\phi(x)dx \\
&= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} \phi(x) \exp(-ixy)dx \right] f(y)dy \\
&= \langle T_f, F\phi \rangle
\end{align*}
Consider again the Dirac delta $\delta$ defined above.
\begin{align*}
\langle F\delta, \phi \rangle &= \langle \delta , F\phi \rangle \\
&= (F\phi)(0) \\
&= \int_{\mathbb{R}} \phi(x) dx \\
&= \int_{\mathbb{R}} 1 \cdot \phi(x) dx \\
&= \langle T_1, \phi \rangle
\end{align*}
So, the Fourier transform of $\delta$ is the linear functional acting on the space $S(\mathbb{R})$ induced by the constant, $1$. Now, here's the cool part that we can't do by treating things in the "normal" way. We can find the Fourier transform of 1:
\begin{align*}
\langle FT_1, \phi \rangle &= \langle T_1, F\phi \rangle \\
&= \int_{\mathbb{R}} (F\phi)(x) dx \\
&= 2\pi \frac{1}{2\pi} \int_{\mathbb{R}} (F\phi)(x) \exp(ix\cdot 0) dx \\
&= 2\pi \phi(0) \\
&= \langle 2\pi\delta, \phi \rangle \\
\end{align*}
So $FT_1 = 2\pi \delta$.
