Statement -1 : When treated with `AgNO_(2)` ethyl bromide gives `CH_(3) CH_(2)-NO_(2)` as the major product
Statement -2 : `overset(Theta)(NO_(2))` is an ambident nucleophile

Answer

To solve the question, we need to analyze both statements provided and verify their correctness based on our knowledge of organic chemistry, specifically regarding nucleophilic substitution reactions and the nature of ambident nucleophiles.

### Step-by-Step Solution:

**Step 1: Analyze Statement 1**
- The first statement claims that when ethyl bromide (CH₃CH₂Br) is treated with AgNO₂, it produces nitroethane (CH₃CH₂NO₂) as the major product.
- Ethyl bromide is a primary haloalkane, and when it reacts with AgNO₂, the bromide ion (Br⁻) is replaced by the nitrite ion (NO₂⁻) through a nucleophilic substitution reaction.

**Step 2: Mechanism of Reaction**
- In this reaction, the bromide ion acts as a leaving group. The AgNO₂ dissociates to give Ag⁺ and NO₂⁻.
- The NO₂⁻ ion, being a nucleophile, attacks the electron-deficient carbon of ethyl bromide. This leads to the formation of nitroethane (CH₃CH₂NO₂) and AgBr as a by-product.

**Step 3: Conclusion for Statement 1**
- Since the reaction indeed produces nitroethane as the major product, Statement 1 is correct.

**Step 4: Analyze Statement 2**
- The second statement claims that NO₂⁻ is an ambident nucleophile.
- An ambident nucleophile is one that has two different atoms capable of donating a pair of electrons to an electrophile.

**Step 5: Nature of NO₂⁻**
- The nitrite ion (NO₂⁻) has two potential sites for nucleophilic attack: the nitrogen atom (N) and one of the oxygen atoms (O).
- This means that NO₂⁻ can attack electrophiles from either the nitrogen side or the oxygen side, confirming its ambident nature.

**Step 6: Conclusion for Statement 2**
- Since NO₂⁻ can act as a nucleophile from two different sites, Statement 2 is also correct.

### Final Conclusion
Both statements are correct. Therefore, the answer to the question is that both Statement 1 and Statement 2 are true.

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