a) Find the Thevenin equivalent with respect to terminals a and b by finding the open-circuit voltage and the short-circuit current. (Items in Blue were added to assist in the solution.)

The voltage source directly connects node 1 to the reference node, thus there will be no KCL equation needed at node 1, and

V₁ = 17.4 V

Next write KCL at node 3.

Summing the current out of node 3 gives

(V₃ - 17.4) / 26 + V₃ / 14 - 0.1 = 0

Multiplying through by 182 and simplifying gives

7 V₃ - 121.8 + 13 V₃ - 18.2 = 0
20 V₃ = 140
V₃ = 7 V

Note that at this point, we actually do not have to find V₂ since we can apply voltage division to V₃ to obtain V_oc.

V_oc = V₃ [10 / 14]
V_oc = 5 V = V_Th

Note that the 10 ohm resistor is shorted out, so the circuit is equivalent to the following:

Once again,

V₁ = 17.4 V

Summing the current out of node 3 gives

(V₃ - 17.4) / 26 + V₃ / 4 - 0.1 = 0

Multiplying through by 52 and simplifying gives

2 V₃ - 34.8 + 13 V₃ - 5.2 = 0
15 V₃ = 40
V₃ = 2.667 V

Once again, we see that we do not need to determine V₂ in order to find the desired parameter.

I_sc = V₃ / 4 = 666.7 mA

thus

R_Th = V_oc / I_sc = 5 / 666.7m
R_Th = 7.5 W

and the Thevenin equivalent circuit is

Killing the sources gives

but the 40 W and 15 W in series are shorted out, so the circuit is

Hopefully it is not too difficult to see that this is equivalent to 30 W and 10 W in parallel, or

R_Th = 7.5 W

which agrees with the original value determined for R_Th, giving us some confidence in the answer.