Given a real number $a$ and a sequence $(x_n)_{n=1}^\infty$ defined by:
$$\left\{\begin{matrix} x_1=1 \\ x_2=0 \\ x_{n+2}=\frac{x_n^2+x_{n+1}^2}{4}+a\end{matrix}\right.$$
for all positive integers $n$.
1. For $a=0$, prove that $(x_n)$ converges.
2. Determine the largest possible value of $a$ such that $(x_n)$ converges.
### Part 1: For \( a = 0 \), prove that \( (x_n) \) converges.

1. **Initial Conditions and Recurrence Relation:**
   Given the sequence \( (x_n) \) with initial conditions \( x_1 = 1 \) and \( x_2 = 0 \), and the recurrence relation:
   \[
   x_{n+2} = \frac{x_n^2 + x_{n+1}^2}{4}
   \]
   for all positive integers \( n \).

2. **Analyzing the Recurrence Relation:**
   For \( a = 0 \), the recurrence relation simplifies to:
   \[
   x_{n+2} = \frac{x_n^2 + x_{n+1}^2}{4}
   \]
   We need to show that this sequence converges.

3. **Boundedness:**
   We will show that the sequence \( (x_n) \) is bounded. Notice that:
   \[
   x_3 = \frac{x_1^2 + x_2^2}{4} = \frac{1^2 + 0^2}{4} = \frac{1}{4}
   \]
   \[
   x_4 = \frac{x_2^2 + x_3^2}{4} = \frac{0^2 + \left(\frac{1}{4}\right)^2}{4} = \frac{1}{64}
   \]
   We observe that each term is non-negative and appears to be decreasing.

4. **Monotonicity:**
   We will show that \( x_n \) is decreasing. Assume \( x_n \geq x_{n+1} \) for some \( n \). Then:
   \[
   x_{n+2} = \frac{x_n^2 + x_{n+1}^2}{4} \leq \frac{x_n^2 + x_n^2}{4} = \frac{2x_n^2}{4} = \frac{x_n^2}{2}
   \]
   Since \( x_n \geq x_{n+1} \), it follows that \( x_{n+2} \leq x_n \).

5. **Convergence:**
   Since \( (x_n) \) is bounded below by 0 and is decreasing, by the Monotone Convergence Theorem, \( (x_n) \) converges.

\[
\blacksquare
\]

### Part 2: Determine the largest possible value of \( a \) such that \( (x_n) \) converges.

1. **Recurrence Relation with \( a \):**
   The recurrence relation is:
   \[
   x_{n+2} = \frac{x_n^2 + x_{n+1}^2}{4} + a
   \]

2. **Fixed Point Analysis:**
   Assume the sequence converges to a limit \( L \). Then:
   \[
   L = \frac{L^2 + L^2}{4} + a = \frac{2L^2}{4} + a = \frac{L^2}{2} + a
   \]
   Solving for \( L \):
   \[
   L = \frac{L^2}{2} + a
   \]
   \[
   2L = L^2 + 2a
   \]
   \[
   L^2 - 2L + 2a = 0
   \]

3. **Discriminant Analysis:**
   For \( L \) to be real, the discriminant of the quadratic equation must be non-negative:
   \[
   \Delta = 4 - 8a \geq 0
   \]
   \[
   4 \geq 8a
   \]
   \[
   a \leq \frac{1}{2}
   \]

4. **Conclusion:**
   The largest possible value of \( a \) such that \( (x_n) \) converges is \( \frac{1}{2} \).

\[
\boxed{\frac{1}{2}}
\]