4. $\tan 37.5^{\circ}=$
4. $\sqrt{6}-2+\sqrt{3}-\sqrt{2}$.

As shown in Figure 2, construct an isosceles right triangle $\triangle ABC$ such that
$$
\begin{array}{l}
\angle C=90^{\circ}, \\
A C=B C=1 .
\end{array}
$$

Then $A B=\sqrt{2}$.
Construct $\angle CAD=30^{\circ}$.
Then $C D=\frac{\sqrt{3}}{3}$,
$$
A D=\frac{2 \sqrt{3}}{3} \text {. }
$$

Construct the angle bisector $AE$ of $\angle BAD$. Let $CE=x$. Then
$$
\begin{array}{l}
B E=1-x, D E=x-\frac{\sqrt{3}}{3} . \\
\text { Hence } \frac{x-\frac{\sqrt{3}}{3}}{\frac{2 \sqrt{3}}{3}}=\frac{1-x}{\sqrt{2}} \\
\Rightarrow x=\frac{\sqrt{2}+1}{\sqrt{3}+\sqrt{2}}=\sqrt{6}-2+\sqrt{3}-\sqrt{2} .
\end{array}
$$

Therefore, $\tan 37.5^{\circ}=\frac{CE}{AC}=\frac{x}{1}=\sqrt{6}-2+\sqrt{3}-\sqrt{2}$.