Memorize resistors in parallel have the same voltage, resistors in series have the same current. Since they have the same voltage, V1 = i1R1 = V2 = i2R2 where R2 = 5R1

i1*R1 = i2*5*R1 SO i1 = 5*i2 and i1/i2 = 5. Can't find i1 or i2 without a voltage.

R2 carries 1/6 the current and R1 carries 5/6 the current (the total resistance is 6 whatevers)

Ratio of currents is 5:1, same as resistance ratio, where 1part is what is flowing in R2.

The ratio of I1 and I2 is simple

For a parallel circuit the same voltage is across both resistor

I_1 =V_T/ ((1/5) *R_2)

I_2 = V_T/R_2

I_1/I_2 = (V_T/ (5*R_2)) / (V_T/R_2) = (R_2/R_2)*(V_T/V_T)*(5 / 1) = 5

so I_1 is 5 times larger than I_2

For 1st question in terms of what

with your information we can only determine the current in terms

of the total current , the voltage of the batttery , etc.

You need to clarify what you want.

so I_1 = ( 5/6) I_T

I_2 = (1/6) I_T

we could use V_T and R_2 to determine a value , what do you want.

If R2 has five times the resistance of R1, then I1 will be 5 times I2, or I2 = 0.2 I1.

Because the voltage "drop" through both branches must be the same.