Analysis
This question mainly examines the application of the sine rule. It tests the students' ability to analyze and solve problems.

By using the sine rule and the given values of $b$ and $\sin B$, we can find the relationship between $a$ and $\sin A$, and use $B$ to find $A+C$. To have two solutions for the triangle, these two values must be complementary. First, if $A \leqslant 45^{\circ}$, then the angle complementary to $A$ would be greater than $135^{\circ}$, leading to a contradiction with the sum of angles in a triangle being $180^{\circ}$. Thus, we can deduce that $45^{\circ} < A < 135^{\circ}$. If $A=90^{\circ}$, its complementary angle is also $90^{\circ}$, which does not meet the condition of having two solutions. From this, we can deduce the range of $\sin A$, and use the relationship between $\sin A$ and $a$ to find the range of $a$.

Solution

Given: $\dfrac {a}{\sin A}= \dfrac {b}{\sin B}=2 \sqrt {2}$
$\therefore a=2\sqrt {2}\sin A$
$A+C=180^{\circ}-45^{\circ}=135^{\circ}$
If $A$ has two values, then these two values are complementary.
If $A\leqslant 45^{\circ}$
Then the angle complementary to $A$ would be greater than or equal to $135^{\circ}$
This would imply $A+B\geqslant 180^{\circ}$, which is not possible.
$\therefore 45^{\circ} < A < 135^{\circ}$
Moreover, if $A=90^{\circ}$, its complementary angle is also $90^{\circ}$, which corresponds to only one solution.
Therefore, $\dfrac { \sqrt {2}}{2} < \sin A < 1$
$a=2\sqrt {2}\sin A$
Thus, $2 < a < 2\sqrt {2}$
Hence, the answer is $\boxed{(2,2\sqrt {2})}$.