Hello, ncbabe!
Here's some help . . .
Code:
(4)
C
* * *
* 3 *
A*    *    *B
* * D 12 * *
*  *R
* *  * *
* * *
O
We have: .$\displaystyle OA = OB = OC = R\text{ (radius)}$
. . . .and: .$\displaystyle AD = DB = 12$
Since $\displaystyle CD = 3$, then: .$\displaystyle DO = R3$
In right triangle BDO, we have: .$\displaystyle (R3)^2 + 12^2 \:=\:R^2$
. . which simplifies to: .$\displaystyle 6R \:=\:153 \quad\Rightarrow\quad\boxed{ R \:=\:\frac{51}{2}}$
Code:
(5)
C
* * *
*  *
*  *
*  *
P
A*      o    *B
* 12.8  2.8 *
*  *

* 22.4  *
*  *
*  *
* * *
D
We're expected to know this theorem.
With two intersecting chords, the product of the segments of one chord
. . equals the product of the segments of the other chord.
In this problem: .$\displaystyle (CP)(PD) \:=\AP)(PB)$
So we have: .$\displaystyle (CP)(22.4) \:=\12.8)(2.8)\quad\Rightarrow\quad\boxed{CP \:=\:1.6} $