Answer:
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω
Explanation:
Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.
When the current increases by 0.450 A wen only R1 is in the circuit, the current is
9/R1 + R2 + 0.450 A = 9/R1       (1)
When the current increases by 0.225 A when only R2 is in the circuit, the current is
9/R1 + R2 + 0.225 A = 9/R2       (2)
equation (1) - (2) equals 
9(1/R1 - 1/R2) = 0.450 A - 0.225 
9(1/R1 - 1/R2) = 0.125 
(1/R1 - 1/R2) = 0.125 A/9 = 0.0138 
1/R1 = 0.0138 + 1/R2
R1 = R2/(1 + 0.0138R2)     (3)
From (1)
9/R1 - 9/R1 + R2 = 0.450 A
9R2/[R1(R1 + R2)] = 0.450 A
R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5 
R2/[R1(R1 + R2)] = 0.5    (4)
From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,
(1 + 0.0138R2) = 0.5(R1 + R2)
0.5R1 + 0.5R2 = 1 + 0.0138R2
0.5R1 = 1 + 0.0138R2 - 0.5R2
0.5R1 = 1 - 0.4862R2        (5)
Substituting (3) into (5) we have
0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2
R2 = (1 + 0.0138R2)(1 - 0.4862R2)
R2 = 1 - 0.4724R2 - 0.0067R2²
Collecting like terms, we have
0.0067R2² + 0.4724R2 + R2 - 1 = 0
0.0067R2² + 1.4724R2 - 1 = 0
Using the quadratic formula, 

We choose the positive answer.
So R2 = 0.340 Ω
From (5)
R1 = 0.5 - 0.9931R2 
    = 0.5 - 0.9931 × 0.340
    = 0.5 - 0.338 
    = 0.162 Ω
a. R1 = 0.162 Ω
b. R2 = 0.340 Ω