Given that $x = 4$, let's solve each part of the question step-by-step.

Part (a): Analyzing the Circuit

We need to determine the voltage at the inverting, non-inverting terminals, and the output of $OA-1$.

1. Non-inverting terminal (+) of OA-1:

   • The non-inverting terminal is connected to a voltage divider network consisting of resistors $(1+x) \, k\Omega$ and $1 \, k\Omega$.
   • Since $x = 4$, $(1+x) \, k\Omega = (1+4) \, k\Omega = 5 \, k\Omega$.
   • Applying the voltage divider rule: $V_{+} = \frac{5 \, k\Omega}{5 \, k\Omega + 1 \, k\Omega} \times 5 \, V = \frac{5}{6} \times 5 \, V = \frac{25}{6} \, V \approx 4.17 \, V$

2. Inverting terminal (-) of OA-1:

   • Since this is an ideal operational amplifier in a negative feedback configuration, the voltage at the inverting terminal will be the same as the non-inverting terminal due to the virtual short concept.
   • Therefore, $V_{-} = V_{+} = 4.17 \, V$.

3. Output of OA-1:

   • The output voltage $V_{out1}$ of $OA-1$ will be such that it maintains the voltage at the inverting terminal equal to the non-inverting terminal.
   • Using Ohm's law across the $1 \, M\Omega$ resistor, we calculate $V_{out1}$ by recognizing that $1 \, \text{M}\Omega$ resistor carries the current from $V_{out1}$ to $V_{-}$ (4.17 V).

   Since the inverting input is grounded via $V_{out1}$, and there’s a $1 \, M\Omega$ resistor: [