Given P(W₁) = (1)/(4).

Step 1: Define the probabilities for the first game.
Let $P(W_1)$ be the probability of winning the first game.
Let $P(L_1)$ be the probability of losing the first game.
Given $P(W_1) = \frac{1}{4}$.
The probability of losing the first game is $P(L_1) = 1 - P(W_1) = 1 - \frac{1}{4} = \frac{3}{4}$.

Step 2: Determine the conditional probabilities for winning the second game.
Let $P(W_2 | W_1)$ be the probability of winning the second game given that the first game was won.
"If it wins, the probability of winning the next game increases by 10%." This means the probability value itself increases by 0.10.

$$P(W_2 | W_1) = P(W_1) + 0.10 = \frac{1}{4} + 0.10 = 0.25 + 0.10 = 0.35$$

Let $P(W_2 | L_1)$ be the probability of winning the second game given that the first game was lost.
"If it loses, the probability of winning the next game increases by 20%." This means the probability value itself increases by 0.20.

$$P(W_2 | L_1) = P(W_1) + 0.20 = \frac{1}{4} + 0.20 = 0.25 + 0.20 = 0.45$$

Step 3: Calculate the probability that the team wins in the second round, $P(W_2)$, using the law of total probability.

$$P(W_2) = P(W_2 | W_1) P(W_1) + P(W_2 | L_1) P(L_1)$$

Substitute the values:

$$P(W_2) = (0.35) \times \left(\frac{1}{4}\right) + (0.45) \times \left(\frac{3}{4}\right)$$

Convert the decimals to fractions: $0.35 = \frac{35}{100} = \frac{7}{20}$ and $0.45 = \frac{45}{100} = \frac{9}{20}$.

$$$$