Op-Amp and resistor network problem

Hi
First, because the stimulus I1 is a DC value, I'll conveniently ignore the effect of C1.

V(n002) is then I1*(R2+[R1|R3])
In fact, it is approximately true we can ignore R1 and R3 then we have
V(n002) ~= I1*R2 = 1.144 V

This is so because of the 'golden rules': The op amp feedback keeps V- at zero, and essentially no current flows into the input pins.

Because R3<<R1, it is not worth finding the exact value of this parallel network, it will be 10 Ohms to within a very close tolerance - but also 10R << 71500 so the feedback at DC is about the same as R2 alone.
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Now for V(n007).

First, ignore R10 because it is directly across V1 and V1 does not care.
Then we can find the "Thevenin equivalent" circuit for V1 feeding R5 & R6, call this Veq
This is equivalent to a voltage Veq=(B.V1) with series resistance R5|R6.
It looks like this

GND---[Veq---Req]---v(n007)

Veq = 1.25*(215/(215+8.87))=1.20
Req = 8.52k

B is found from the potential divider equation: R6/(R6+R5) = 0.96
The equivent series resistance is R5|R6 because an ideal voltage source has no resistance. Thus current in or out of the divider R6/(R6+R5) sees a resistance R5|R6.
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Now we tie n002 to n007 using R7:

GND---v(n007)---R7---x---Req---Veq---GND
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n007 = 1.144 + (1.2-1.144)*(R7/(R7+Req))
= 1.144 + 0.056*0.34
= 1.16304