2^(a+2) = (2^a)*(2^2) = 4*(2^a)
3^(b+1) = (3^b)*(3^1) = 3*(3^b)

Hence, 2^(a+2) - 3^(b+1) = 4*(2^a) - 3*(3^b) = 5 ............................. (A)
Multiplying the other equation by 3, 3*(2^a) + 3*(3^b) = 3*17 = 51 ........... (B)
Adding (A) and (B), 7*(2^a) = (51 + 5) = 56 ---> (2^a) = 8 = 2^3 ---> a = 3
Now, (3^b) = 17 - (2^3) = 17 - 8 = 9 = 3^2 ---> b = 2

rattan123 wrote:change in the question
2^a + 3^b = 17
2^a+b -3^b+1 = 5
Find the value of a and b.

From the first equation, (3^b) = 17 - (2^a)
Now, 3^(b+1) = (3^b)*(3^1) = 3*(3^b) = 3*(17 - (2^a)) = 51 - 3*(2^a)
And, 2^(a+b) = (2^a)*(2^b)

--> 2^(a+b) - 3^(b+1) = 5
--> (2^b)*(2^a) - [51 - 3*(2^a)] = 5
--> (2^b)*(2^a) + 3*(2^a) = 56
--> (2^a)[(2^b) + 3] = 56

Now unless it is mentioned a and b are positive integers, it is beyond the scope of GMAT syllabus to find the value of a and b from this.

If we assume a and b are positive integers, then both (2^a) and (3^b) will be integer. Now, 56 can expressed as a product of two integers in following ways : 1*56, 2*28, 4*14, 7*8

Only 7*8 can be written in the form (2^a)[(2^b) + 3] for a = 3 and b = 2.