7. In the tetrahedron $S-ABC$, it is known that
$$
\begin{array}{l}
\angle SAB = \angle SAC = \angle ACB = 90^{\circ}, \\
AC = 2, BC = \sqrt{13}, SB = \sqrt{29}.
\end{array}
$$

Then the cosine of the angle formed by the lines $SC$ and $AB$ is $\qquad$
7. $\frac{\sqrt{17}}{17}$.

As shown in Figure 4, take $A$ as the origin, and the lines $A B$ and $A S$ as the $y$-axis and $z$-axis, respectively, to establish a spatial rectangular coordinate system.
Then the points are $B(0, \sqrt{17}, 0), S(0,0,2 \sqrt{3})$,
$C\left(2 \sqrt{\frac{13}{17}}, \frac{4}{\sqrt{17}}, 0\right)$.
Thus, $\overrightarrow{S C}=\left(2 \sqrt{\frac{13}{17}}, \frac{4}{\sqrt{17}},-2 \sqrt{3}\right)$,
$$
\overrightarrow{A B}=(0, \sqrt{17}, 0) \text {. }
$$

Therefore, the cosine value of the required angle is
$$
\frac{\overrightarrow{S C} \cdot \overrightarrow{A B}}{|\overrightarrow{S C}||\overrightarrow{A B}|}=\frac{\sqrt{17}}{17} .
$$