### Problem Solution Rewritten in Step-by-Step Format

#### Part (1): Latest Time Xiao Li Should Leave Home

- Case of Leaving at 7:46 a.m.: Xiao Li will definitely not be late because the journey without red lights takes 10 minutes, arriving at 7:56 a.m.
- Case of Leaving at 7:47 a.m.: Xiao Li will be late only if he encounters 4 red lights. The probability of this happening is calculated as:
$$P(\text{4 red lights}) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$
Thus, the probability of not being late is:
$$P(\text{not late}) = 1 - \frac{1}{16} = \frac{15}{16} > 90\%$$
- Case of Leaving at 7:48 a.m.: Xiao Li will be late if he encounters 3 or 4 red lights. The probability of being late is calculated as:
$$P(\text{3 or 4 red lights}) = {C}_{4}^{3} \times \left(\frac{1}{2}\right)^3 \times \frac{1}{2} + \left(\frac{1}{2}\right)^4 = \frac{5}{16}$$
Therefore, the probability of not being late is:
$$P(\text{not late}) = 1 - \frac{5}{16} = \frac{11}{16} < 90\%$$
To ensure a probability of not being late higher than 90%, Xiao Li should leave home at $\boxed{7:47\, \text{a.m.}}$

#### Part (2): Probability of Being Late on Exactly One Day

- Given the probability of being late when leaving at 7:48 a.m. is $\frac{5}{16}$, and the probability of not being late is $\frac{11}{16}$.
- The probability of being late on exactly one day out of two is calculated using the binomial formula:
$$P = {C}_{2}^{1} \times \frac{5}{16} \times \frac{11}{16} = \frac{55}{128}$$
Therefore, the probability is $\boxed{\frac{55}{128}}$.

#### Part (3): Average Duration of Xiao Li's Journey to Work

- The possible journey times and their probabilities are calculated as follows:
- $P(X=10) = P(X=14) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$
- $P(X=11) = P(X=13) = {C}_{4}^{1} \times \left(\frac{1}{2}\right)^4 = \frac{1}{4}$
- $P(X=12) = {C}_{4}^{2} \times \left(\frac{1}{2}\right)^4 = \frac{3}{8}$
- The average duration, $E(X)$, is calculated as:
$$E(X) = 10 \times \frac{1}{16} + 11 \times \frac{1}{4} + 12 \times \frac{3}{8} + 13 \times \frac{1}{4} + 14 \times \frac{1}{16} = 12\, \text{minutes}$$
Therefore, the average duration of Xiao Li's journey to work is $\boxed{12\, \text{minutes}}$.