Why is $f(x) \delta(x) = f(0)\delta(x)$ only true when $x=0$? This is a follow up from a previous question asked by me. 
I know that $$\delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases} $$  and that 
$$\int_{-\infty}^{\infty} \delta(x) \mathrm{d}x = 1$$
I also know that the product $$f(x) \delta(x)=  0\space\forall \space x\ne 0$$ 
I can summarize my lack of understanding with basically two questions: 
$\color{red}{\mathrm{Question} \space1:}$ If I'm allowed to write $f(x) \delta(x) = f(0)\delta(x)$ then why not $f(x) \delta(x) = f(3)\delta(x)$ since $3 \ne 0$?
$\color{blue}{\mathrm{Question} \space2:}$ Also, why do we just substitute $x=0$ into the function $f(x)$ and not $\delta(x)$? In other words why don't we write $f(0)\delta(0)$ or $f(3)\delta(3)$ instead of $f(0)\delta(x)$ and $f(3)\delta(x)$ respectively. I know that $f(0)\delta(0)$ is undefined, but the point is that the $\delta$ still takes $x$ as its argument as well as $f$.
(As ever, apologies for the abuse of notation; this Dirac measure is all very new to me, hence all the questions about it) 


PRIMER:
In This Answer and This Answer, I provided more detailed primers on the Dirac Delta.  Herein, we condense the content of those answers.
The Dirac is not a function, but rather a Generalized Function also known as a Distribution.  
The symbol $\int_{-\infty}^{\infty}\delta (x)f(x)\,dx$ is ,in fact, not an integral.  It is a Functional that maps a test function $f$ into the number given by $f(0)$. (Note that whereas a function is a mapping, or "rule" that assigns to a number in a domain, a number in the range, a functional is a "rule" that assigns to functions in a vector space domain, a number.
We write
$$\int_{-\infty}^{\infty}\delta (x-a)f(x)\,dx=f(a)$$
but alternatively, and more compactly, we can write
$$\langle \delta_a,f \rangle=f(a)$$
For $a=0$, we have
$$\int_{-\infty}^{\infty}\delta(x)f(x)\,dx=\langle \delta_0,f \rangle=f(0)$$

Now, for this specific question, we have 
$$\begin{align}
\int_{\infty}^{\infty}f(x)\delta(x)\,dx&=f(0)\\\\
&=\int_{-\infty}^{\infty}f(0)\delta(x)\,dx\\\\
&\ne f(3)\\\\
&=\int_{-\infty}^{\infty}f(3)\delta(x)\,dx\end{align}$$
Therefore, $f(x)\delta(x)=f(0)\delta(x)\ne f(3)\delta(x)$ as was to be shown.
