7. Given $|\boldsymbol{a}|=1,|\boldsymbol{b}|=|c|=2, \boldsymbol{b} \cdot \boldsymbol{c}=0$, $\lambda \in(0,1)$. Then
$$
|a-b+\lambda(b-c)|+\left|\frac{1}{2} c+(1-\lambda)(b-c)\right|
$$

the minimum value is $\qquad$
7. $\sqrt{5}-1$.

By the property of rotation, without loss of generality, let
$$
a=(\cos \theta, \sin \theta), b=(2,0), c=(0,2) .
$$

Then $|a-b+\lambda(b-c)|+\left|\frac{1}{2} c+(1-\lambda)(b-c)\right|$
$$
\begin{aligned}
= & \sqrt{(\cos \theta-2+2 \lambda)^{2}+(\sin \theta-2 \lambda)^{2}}+ \\
& \sqrt{(2-2 \lambda)^{2}+(2 \lambda-1)^{2}} .
\end{aligned}
$$

Let point $A(\cos \theta, \sin \theta), B(2-2 \lambda, 2 \lambda)$, $C(0,1)$. Then point $A$ is on the unit circle, point $B$ is on the line $x+y=2$, and construct the symmetric point $C^{\prime}(2,1)$ of point $C$ with respect to $x+y=2$.
Therefore, the value we are looking for is
$$
\begin{array}{l}
|A B|+|B C|=|A B|+\left|B C^{\prime}\right| \\
\geqslant\left|A C^{\prime}\right| \geqslant\left|O C^{\prime}\right|-1=\sqrt{5}-1 .
\end{array}
$$