Rewrite the expression in terms of sine and cosine.

The given limit is $\lim_{x \to 1/2} \frac{\tan 3\pi x}{\sec \pi x}$.
We know that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$.
So, the expression becomes:

Substitute $x = 1/2$ into the rewritten expression:
Numerator: $\sin\left(3\pi \cdot \frac{1}{2}\right) \cdot \cos\left(\pi \cdot \frac{1}{2}\right) = \sin\left(\frac{3\pi}{2}\right) \cdot \cos\left(\frac{\pi}{2}\right) = (-1) \cdot (0) = 0$.
Denominator: $\cos\left(3\pi \cdot \frac{1}{2}\right) = \cos\left(\frac{3\pi}{2}\right) = 0$.
Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule.

L'Hôpital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$.
Let $f(x) = \sin 3\pi x \cdot \cos \pi x$ and $g(x) = \cos 3\pi x$.

Find the derivative of $f(x)$ using the product rule $(uv)' = u'v + uv'$:
$f'(x) = \frac{d}{dx}(\sin 3\pi x \cdot \cos \pi x)$
$f'(x) = (3\pi \cos 3\pi x)(\cos \pi x) + (\sin 3\pi x)(-\pi \sin \pi x)$
$f'(x) = 3\pi \cos 3\pi x \cos \pi x - \pi \sin 3\pi x \sin \pi x$.

Find the derivative of $g(x)$:
$g'(x) = \frac{d}{dx}(\cos 3\pi x)$
$g'(x) = -3\pi \sin 3\pi x$.

Substitute $x = 1/2$ into $f'(x)$:
$f'\left(\frac{1}{2}\right) = 3\pi \cos\left(\frac{3\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - \pi \sin\left(\frac{3\pi}{2}\right) \sin\left(\frac{\pi}{2}\right)$
$f'\left(\frac{1}{2}\right) = 3\pi (0)(0) - \pi (-1)(1)$
$f'\left(\frac{1}{2}\right) = 0 - (-\pi) = \pi$.

Substitute $x = 1/2$ into $g'(x)$:
$g'\left(\frac{1}{2}\right) = -3\pi \sin\left(\frac{3\pi}{2}\right)$
$g'\left(\frac{1}{2}\right) = -3\pi (-1) = 3\pi$.

Now, calculate the limit:

The value of the limit is $\frac{1}{3}$. This corresponds to option b.

The final answer is $\text{b. } \frac{1}{3}$.