If the function $\mathrm{f}(x)=\left\{\begin{array}{ccc}x+a \sqrt{2} \sin x & \text { if } & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+\mathrm{b} & \text { if } & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-\mathrm{bsin} x & \text { if } & \frac{\pi}{2} < x \leq \pi\end{array}\right.$ is continuous in $[0, \pi]$ then $a-b=$

Continuity requires the function to match at transition points $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.

At $x = \frac{\pi}{4}$, the left limit equals the function value: $\lim_{x \to \frac{\pi}{4}^-} f(x) = f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + a$.
The right limit is $\lim_{x \to \frac{\pi}{4}^+} f(x) = \frac{\pi}{2} + b$.
Equating gives $\frac{\pi}{4} + a = \frac{\pi}{2} + b$, yielding $a - b = \frac{\pi}{4}$.

At $x = \frac{\pi}{2}$, the function value is $f\left(\frac{\pi}{2}\right) = b$.
The right limit is $\lim_{x \to \frac{\pi}{2}^+} f(x) = -a - b$.
Equating gives $b = -a - b$, or $a + 2b = 0$.

Solving the system $a - b = \frac{\pi}{4}$ and $a + 2b = 0$:

From the second equation, $a = -2b$. Substituting into the first: $-2b - b = \frac{\pi}{4}$, so $-3b = \frac{\pi}{4}$ and $b = -\frac{\pi}{12}$.
Then $a = -2\left(-\frac{\pi}{12}\right) = \frac{\pi}{6}$.

The difference is $a - b = \frac{\pi}{6} - \left(-\frac{\pi}{12}\right) = \frac{\pi}{4}$.

$\boxed{\frac{\pi}{4}}$