The area of right triangle $ABC$ is 4, and the hypotenuse $\overline{AB}$ is 12.  Compute $\sin 2A.$
We have that $\frac{1}{2} ab = 4,$ so $ab = 8.$

[asy]
unitsize (1 cm);

pair A, B, C;

C = (0,0);
B = (3,0);
A = (0,2);

draw(A--B--C--cycle);
draw(rightanglemark(A,C,B,6));

label("$A$", A, N);
label("$B$", B, E);
label("$C$", C, SW);
label("$a$", (B + C)/2, S, red);
label("$b$", (A + C)/2, W, red);
label("$12$", (A + B)/2, NE, red);
[/asy]

Then
\[\sin 2A = 2 \sin A \cos A = 2 \cdot \frac{a}{12} \cdot \frac{b}{12} = \frac{ab}{72} = \frac{8}{72} = \boxed{\frac{1}{9}}.\]