Fundamental period of periodic delta function
Okay let's go with the definition of periodicity:

\begin{equation*}
x[n]=x[n+T]
\end{equation*}

which in turn yields:

\begin{eqnarray*}
x[n+T] = \sum\limits_{k=-\infty}^{\infty} (\delta[n+T-4k]+\delta[n+T-4k-1]-\delta[n+T-4k-2]-\delta[n+T-4k-3]) \\ = \sum\limits_{k=-\infty}^{\infty} (\delta[n-(4k-T)]+\delta[n-(4k+1-T]-\delta[n-(4k+2-T)]-\delta[n-(4k+3-T)])
\end{eqnarray*}

On inspection, if we allow T=4, 

\begin{eqnarray*}
x[n+4] = \sum\limits_{k=-\infty}^{\infty} (\delta[n-(4k-4)]+\delta[n-(4k+1-4]-\delta[n-(4k+2-4)]-\delta[n-(4k+3-4)]) \\ = \sum\limits_{k=-\infty}^{\infty} (\delta[n-4(k-1)]+\delta[n-(4k-3)]-\delta[n-(4k-2)]-\delta[n-(4k-1)])
\end{eqnarray*}

Now let k-1=m or m=k+1. Note that limits on m remain the same:

\begin{eqnarray*}
x[n+4] = \sum\limits_{m=-\infty}^{\infty} (\delta[n-4m]+\delta[n-4m-1)]-\delta[n-4m-2]-\delta[n-4m-3)]) \\ = x[n]
\end{eqnarray*}

This explains why the period of x[n] is 4. Naturally, the period is directly related to the coefficient of k, as can be seen in the above equations. 

As for x[0], n takes the value of 0, but k continues to range from -infinity to infinity. 

\begin{equation*}
x[0] = \sum\limits_{k=-\infty}^{\infty} (\delta[-4k]+\delta[-4k-1]-\delta[-4k-2]-\delta[-4k-3])
\end{equation*}