Statement 1: The circumcircle of a triangle formed by the lines `x=0,x+y+1=0` and `x-y+1=0` also passes through the point (1, 0). Statement 2: The circumcircle of a triangle formed by three tangents of a parabola passes through its focus.

Answer

To solve the problem, we will analyze both statements one by one.

### Step 1: Analyze Statement 1
We need to determine if the circumcircle of the triangle formed by the lines \( x = 0 \), \( x + y + 1 = 0 \), and \( x - y + 1 = 0 \) passes through the point \( (1, 0) \).

1. **Find the points of intersection of the lines:**
- The line \( x = 0 \) intersects \( x + y + 1 = 0 \):
\[ x = 0 \implies 0 + y + 1 = 0 \implies y = -1 \implies A(0, -1) \]
- The line \( x = 0 \) intersects \( x - y + 1 = 0 \):
\[ x = 0 \implies 0 - y + 1 = 0 \implies y = 1 \implies B(0, 1) \]
- The lines \( x + y + 1 = 0 \) and \( x - y + 1 = 0 \) intersect:
\[ \begin{align*} x + y + 1 &= 0 \quad (1) \\ x - y + 1 &= 0 \quad (2) \end{align*} \]
Adding (1) and (2):
\[ 2x + 2 = 0 \implies x = -1 \]
Substituting \( x = -1 \) into (1):
\[ -1 + y + 1 = 0 \implies y = 0 \implies C(-1, 0) \]

2. **Points of the triangle:**
The vertices of the triangle are \( A(0, -1) \), \( B(0, 1) \), and \( C(-1, 0) \).

### Step 2: Find the circumcircle
The circumcircle of a triangle can be represented by the general equation:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
We will substitute the coordinates of points A, B, and C to find \( g, f, c \).

1. **Substituting point A(0, -1):**
\[ 0^2 + (-1)^2 + 2g(0) + 2f(-1) + c = 0 \implies 1 - 2f + c = 0 \quad (1) \]

2. **Substituting point B(0, 1):**
\[ 0^2 + 1^2 + 2g(0) + 2f(1) + c = 0 \implies 1 + 2f + c = 0 \quad (2) \]

3. **Substituting point C(-1, 0):**
\[ (-1)^2 + 0^2 + 2g(-1) + 2f(0) + c = 0 \implies 1 - 2g + c = 0 \quad (3) \]

### Step 3: Solve the equations
From equations (1), (2), and (3):
1. From (1): \( c = 2f - 1 \)
2. From (2): \( c = -2f - 1 \)
Setting these equal:
\[ 2f - 1 = -2f - 1 \implies 4f = 0 \implies f = 0 \]
Substituting \( f = 0 \) into \( c = 2f - 1 \):
\[ c = -1 \]
3. Substituting \( f = 0 \) into (3):
\[ 1 - 2g - 1 = 0 \implies g = 0 \]

### Step 4: Equation of the circumcircle
The circumcircle is:
\[ x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1 \]
Now we check if the point \( (1, 0) \) lies on this circle:
\[ 1^2 + 0^2 = 1 \implies 1 = 1 \]
Thus, the point \( (1, 0) \) lies on the circumcircle.

### Conclusion for Statement 1
Statement 1 is **True**.

### Step 5: Analyze Statement 2
The statement claims that the circumcircle of a triangle formed by three tangents of a parabola passes through its focus. This is a known property of conics, specifically parabolas, and is **True**.

### Final Answer
Both statements are **True**.