for the 1 problem the lower left box is i1 the lower right side is i2 the top part is i3
the method i used is called mesh analysis which is basically KCL
here is the 3 equations i got for each current
[i1] 4mA (i get a direct answer instead of an equation for this current part because there is a current source on the box i am evaluating which makes i1=4mA
[i2] 8(i2-i1) + 20 +10i2
[i3] -20 -2(i3-i1)+5i3
now you have i1=4mA which simplifes i2 and i3 equations now you have two unkowns and two variables elinimate one current and solve for one and plug it in to solve for the other one.
for number 2
you have three equations for this as well they will all be on the top part of the circuit V1 is the top left corner between the 10V and the 4 ohms resisitor, V2 is between the 4 ohms resistor and the 2 ohms resistor and V3 is the point where 2Vo and 5A meet.
[V1] 10V ( you are trying to find the voltages at these points 10V goes directly into the V1 node which we assigned so yea)
[V2] V2-V1/4 + V2/10 + V2-V3/2+2Vo
[V3] v3/20+V3-V2/-2Vo +2
please rate