$a,b,c$ - are sides of triangle $T$. It is known, that if we increase any one  side by $1$, we get new
a) triangle
b)acute triangle

Find minimal possible area of triangle $T$ in case of a) and in case b)
To solve the problem, we need to find the minimal possible area of triangle \( T \) under two conditions:
a) Increasing any one side by 1 still results in a triangle.
b) Increasing any one side by 1 results in an acute triangle.

Let's start with condition b) since the provided solution addresses this part.

### Condition b: Acute Triangle

1. **Assume \( a \ge b \ge c \)**:
   - For the triangle to remain acute when any side is increased by 1, the following inequalities must hold:
     \[
     (a+1)^2 \le b^2 + c^2
     \]
     \[
     (b+1)^2 \le a^2 + c^2
     \]
     \[
     (c+1)^2 \le a^2 + b^2
     \]

2. **Simplify the inequalities**:
   - Consider the first inequality:
     \[
     (a+1)^2 \le b^2 + c^2
     \]
     Expanding and simplifying:
     \[
     a^2 + 2a + 1 \le b^2 + c^2
     \]
     \[
     2a + 1 \le b^2 + c^2 - a^2
     \]

   - Consider the second inequality:
     \[
     (b+1)^2 \le a^2 + c^2
     \]
     Expanding and simplifying:
     \[
     b^2 + 2b + 1 \le a^2 + c^2
     \]
     \[
     2b + 1 \le a^2 + c^2 - b^2
     \]

   - Consider the third inequality:
     \[
     (c+1)^2 \le a^2 + b^2
     \]
     Expanding and simplifying:
     \[
     c^2 + 2c + 1 \le a^2 + b^2
     \]
     \[
     2c + 1 \le a^2 + b^2 - c^2
     \]

3. **Find the minimal value of \( c \)**:
   - From the inequality \( 2c + 1 \le c^2 \):
     \[
     c^2 - 2c - 1 \ge 0
     \]
     Solving the quadratic inequality:
     \[
     c \ge 1 + \sqrt{2}
     \]

4. **Verify the sides**:
   - If \( a = b = c = 1 + \sqrt{2} \), we need to check if the conditions hold:
     \[
     (1 + \sqrt{2} + 1)^2 \le (1 + \sqrt{2})^2 + (1 + \sqrt{2})^2
     \]
     \[
     (2 + \sqrt{2})^2 \le 2(1 + \sqrt{2})^2
     \]
     \[
     4 + 4\sqrt{2} + 2 \le 2(3 + 2\sqrt{2})
     \]
     \[
     6 + 4\sqrt{2} \le 6 + 4\sqrt{2}
     \]
     This holds true.

5. **Calculate the area**:
   - The area of an equilateral triangle with side length \( s \) is given by:
     \[
     \text{Area} = \frac{\sqrt{3}}{4} s^2
     \]
     Substituting \( s = 1 + \sqrt{2} \):
     \[
     \text{Area} = \frac{\sqrt{3}}{4} (1 + \sqrt{2})^2
     \]
     \[
     \text{Area} = \frac{\sqrt{3}}{4} (3 + 2\sqrt{2})
     \]

### Conclusion for Condition b:
\[
\boxed{\frac{\sqrt{3}}{4} (3 + 2\sqrt{2})}
\]

### Condition a: Triangle

1. **Triangle Inequality**:
   - For any triangle with sides \( a, b, c \), the triangle inequality must hold:
     \[
     a + b > c
     \]
     \[
     a + c > b
     \]
     \[
     b + c > a
     \]

2. **Increase any side by 1**:
   - If we increase any side by 1, the new sides must still satisfy the triangle inequality:
     \[
     (a+1) + b > c
     \]
     \[
     (b+1) + a > c
     \]
     \[
     (c+1) + a > b
     \]
     \[
     a + (b+1) > c
     \]
     \[
     b + (c+1) > a
     \]
     \[
     c + (a+1) > b
     \]

3. **Minimal sides**:
   - The minimal sides that satisfy these conditions are \( a = b = c = 1 \).

4. **Calculate the area**:
   - The area of an equilateral triangle with side length \( s \) is given by:
     \[
     \text{Area} = \frac{\sqrt{3}}{4} s^2
     \]
     Substituting \( s = 1 \):
     \[
     \text{Area} = \frac{\sqrt{3}}{4} (1)^2
     \]
     \[
     \text{Area} = \frac{\sqrt{3}}{4}
     \]

### Conclusion for Condition a:
\[
\boxed{\frac{\sqrt{3}}{4}}
\]