Question

what part of the total current will flow through a galvanometer of resistance 0.6Ω  when a 0.2Ω shunt is connected to it?

Answer 1

combined resistance:  1/R = 1/0.6  +  1/0.2  = 4/0.6          R = 0.15 ohms
voltage across is same. So current is inversely proportional to resistance.
So  0.15 i /0.6 is the current thru galvanometer.  That is 25%