Why increasing bridge ratio improve efficiency?
$$Bridge\ Ratio = \frac {R_2} {R_1} = \frac {R_3} {R_4} = a$$
You are confusing electrical efficiency vs efficiency of the strain gage.
From p111:

The circuit efficiency, \$\eta\$, of a particular circuit may be expressed as the ratio of its maximum output, per unit of strain, to the corresponding value for the constant-current circuit that produces the maximum obtainable output.

$$\eta = \frac {a} {1+a}\ \ \ (4.36)$$
So if a = 1 (all resistances are equal), the efficiency will be 50%.  a = 10, efficiency = 90.9%, a = 5, 83.3% (above 80%) and a = \$\infty\$, efficiency = 100%.
If \$R_3\$ &gt;&gt; \$R_4\$ (a strain gage), then \$\Delta R_4\$ has smaller impact on current, so essentially constant-current circuit.  Essentially, with a constant-current source, the change in voltage \$\Delta V_4\$ = change in strain.  A Bridge Ratio of 5 is enough to make the results linear.
The higher the bridge ratio, the more linear the circuit.