The Naive Bayes Classifier assumption leads to,
$$P(w_1, w_2, \ldots, w_5 | y) = \prod_{i=1}^{5}P(w_i|y)$$
where $y = \text{Sports/Health}$.
$$P(\text{Sports}|w_1, w_2) = \frac{P(w_1, w_2 | \text{Sports})P(\text{Sports})}{P(w_1, w_2)} = \frac{P(w_1 | \text{Sports})P(w_2 | \text{Sports})P(\text{Sports})}{P(w_1, w_2)}$$
$$P(\text{Health}|w_1, w_2) = \frac{P(w_1, w_2 | \text{Health})P(\text{Health})}{P(w_1, w_2)} = \frac{P(w_1 | \text{Health})P(w_2 | \text{Health})P(\text{Health})}{P(w_1, w_2)}$$
where,
$$\begin{align} P(w_1, w_2) &= P(w_1, w_2 | \text{Sports})P(\text{Sports}) + P(w_1, w_2 | \text{Health})P(\text{Health}) \\\\\\\ &= P(w_1 | \text{Sports})P(w_2 | \text{Sports})P(\text{Sports}) + P(w_1 | \text{Health})P(w_2 | \text{Health})P(\text{Health})\end{align}$$
Take it from here.
**Edit** : Note that,
$$P(y|w_1, w_2, \ldots, w_5) = \frac{P(y)\prod_{i=1}^{5}P(w_i|y)}{P(w_1, w_2, \ldots, w_5)}$$
where $y = \text{Sports/Health}$.