Question

Part A

Answer 1

Lets assume that a current "i" flows in the circuit in counter-clockwise direction.

Using KVL in loop acba

5 i + 1.4 i + 8 + 9 i - 16 + 1.6 i = 0

i = 0.47 A

Part A)

P₅ = Power dissipated in 5 ohm resistor = i² (5) = (0.47)² (5) = 1.1 Watt

Part B)

P₉ = Power dissipated in 9 ohm resistor = i² (9) = (0.47)² (9) = 2.0 Watt

Part B)

P₁₆ = Power output of 16 V battery = i (16) = (0.47) (16) = 7.5 Watt

Answer 2

Since the connection is in series, the current divider can be used. It's mathematical expression is In = (It Rt) / Rn. If the value for is solved, power dissipation can also be solved using P = I^2 R. 
I = 0.47 A 
A.  P = (0.47)^2 (5) = 1.11 Watts
B. P = (0.47)^2 (9) = 2 Watts
C. P = VI = 16(0.47) 4.24 Watts