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*}
the light bulbs are 1.5V @ 0.3A
How do you expect to get 300 mA thru 1 kΩ resistor with only 4.5 V applied? Even with the full 4.5 V across a 1 kΩ resistor, you only get 4.5 mA flowing thru it. With two such resistors in parallel, you get 4.5 mA thru each, so 9 mA total. That's not going to light up a bulb that requires 300 mA for normal brightness. And, this isn't even accounting for the 1.5 V drop across the bulb when it is lit.
One solution is to use lower resistances. Using Ohm's law, we can calculate what they should be. You have 4.5 V available, and the bulb will drop 1.5 V of that when fully lit. That leaves 3.0 V across the resistor. With 3.0 V across it, the resistor should pass 300 mA. (3.0 V)/(300 mA) = 10 Ω. That's the total series resistance for fully lighting the bulb. Since you want that to happen with two resistors in parallel, they should each be 20 Ω.
Also consider the power the resistors will dissipate. Let's be safe and say they need to survive the full 4.5 V applied to them, because stuff inevitably happens. (4.5 V)2/(20 Ω) = 1.0 W. Therefore get "2 W" resistors. Those are also larger and easier to see, so better for demonstration purposes anyway.
Best Answer
The equations are easy to write down, but cumbersome to solve.
When you have a triangle of 3 resistors (a, b, c), and nodes A, B, C. Resistor a is connected between nodes B & C; resistor b between A & C, and c between A & B. Make 3 measurements -- AC, BC, and AB between nodes A&C, B&C and A&B.
Mathematica gives this:
$$a->\frac{(-AB^2+2\cdot AB \cdot AC-AC^2+2\cdot AB\cdot BC+2\cdot AC\cdot BC-BC^2)}{2 \left(AB+AC-BC\right)}$$ $$b->\frac{(-AB^2+2\cdot AB\cdot AC-AC^2+2\cdot AB\cdot BC+2\cdot AC\cdot BC-BC^2)}{2 (AB-AC+BC)}$$ $$c->\frac{(AB^2-2\cdot AB\cdot AC+AC^2-2\cdot AB\cdot BC-2\cdot AC\cdot BC+BC^2)}{2 (AB-AC-BC)}$$