Three. (20 points) Given that $a$ and $b$ are integers, and satisfy $\sqrt{9-8 \sin 50^{\circ}}=a+b \csc 50^{\circ}$. Find all possible values of $a$ and $b$.

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Three, from the triple angle formula, we have
$$
\sin 3 \times 50^{\circ}=3 \sin 50^{\circ}-4 \sin ^{3} 50^{\circ}=\frac{1}{2} \text {. }
$$

Thus, $-8 \sin ^{3} 50^{\circ}+6 \sin 50^{\circ}-1=0$, which means
$\left(9-8 \sin 50^{\circ}\right) \sin ^{2} 50^{\circ}=\left(3 \sin 50^{\circ}-1\right)^{2}$.
Since $9>8 \sin 50^{\circ}$, we have
$$
\sqrt{9-8 \sin 50^{\circ}}=3-\csc 50^{\circ}>0 \text {. }
$$

Therefore, $a=3, b=-1$ satisfies the condition.
Assume there exists another pair $\left(a_{0}, b_{0}\right)$ that satisfies the given condition, then $a_{0} 、 b_{0} \in \mathbf{Z}$, and $a_{0}+b_{0} \csc 50^{\circ}=3-\csc 50^{\circ}$, which means
$$
\left(b_{0}+1\right) \csc 50^{\circ}=3-a_{0} .
$$

Since $b_{0} \neq-1, a_{0} \neq 3$, we have
$$
\sin 50^{\circ}=\frac{b_{0}+1}{3-a_{0}} \text {. }
$$

Thus, $\sin 50^{\circ}$ is a rational number.
Let $\sin 50^{\circ}=\frac{p}{q}$, where $(p, q)=1, p, q \in \mathbf{N}_{+}$. Substituting into
equation (1) and rearranging, we get
$$
8 p^{3}-6 p q^{2}+q^{3}=0 \text {. }
$$

Therefore, $q^{2} \mid 8 p^{3}$.
Since $(p, q)=1$, we know $q^{2} \mid 8$, hence $q \leqslant 2$.
Since $q \neq 1$, we have $q=2$. At this point,
$$
p^{3}-3 p+1=0 \text {. }
$$

By $p\left(p^{2}-3\right)<0$, we have $p=1$, i.e., $\sin 50^{\circ}=\frac{1}{2}$, which is a contradiction.
Therefore, the pair $(a, b)$ that satisfies the given condition is unique.