12-42.

Since this problem doesn't involve springs, we'll have to start from a free body diagram before we can apply the equations we know. Newtons equations will give:

$$\begin{eqnarray*} T\sin(\theta) &=& ma_x\\ T\cos(\theta) &=& mg. \end{eqnarray*}$$

In the $x$ and $y$ directions respectively. Dividing the top equation by the bottom gives

$$\tan(\theta)=a_x/g.$$

Next, since we will only discuss small oscillations (larger oscillations are still harmonic, but are not simple), we will approximate the tangent by sine (remember this procedure - it is often used). Thus we have

$$\frac{a_x}{g}\approx \sin(\theta)=\frac{x}{l},$$

where $l$ is the length of the pendulum. So if we compare this equations with the spring equations from before we see that we have

$$\omega^2=g/l.$$

From this expression we can find $l$ as

$$l=g/\omega^2=.50\;{\textrm m}.$$

To find the maximum kinetic energy we will use the velocity if we remember an equation from circular motion:

$$v=l\frac{d\theta}{dt}=-l\omega(.08\;{\textrm rad})\sin(\omega t + \varphi),$$

so that

$$v_{max}=-l\omega(.08\;{\textrm rad})=.177\;{\textrm m/s},$$

and finally

$$K=94\;{\textrm mJ}.$$