Step 1: Let Puneet's score in test 1 = $a$. Then Urvisha's test 1 = $a$.
Puneet scores same non-zero marks from test 2 to 8, let that be $b$ (b > 0).
Urvisha scores same as Puneet from test 5 onwards: so test 5,6,7,8 = $b$.
Let Urvisha's test 2,3,4 be $c, d, e$ respectively.
Step 2: Puneet's total first 3 tests = $a + b + b = a + 2b$.
Urvisha's total first 2 tests = $a + c$.
Given: $a + 2b = a + c \implies c = 2b$.
Step 3: Urvisha's marks in first test = $a$, total of first 2 tests = $a + c = a + 2b$, total of 8 tests = $a + c + d + e + b + b + b + b = a + 2b + d + e + 4b = a + d + e + 6b$.
These are in GP. So $(a + 2(b)^2 = a \times (a + d + e + 6(b)$.
Step 4: Also, Urvisha's test 3 and 4 ($d, e$) are between 0 and 4, and $b$ is positive integer.
Step 5: Testing optionss: "Urvisha scored 2 marks in the first test" means $a=2$. Then $c=2b$, and $d, e$ are integers. This is possible with $b=1$, $d=3$, $e=4$, etc. So this may be true.
Step 6: Final Answer: Urvisha scored 2 marks in the first test.