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This should work.

1. The colouring:

2. The sequence:

5 R's followed by 5 B's.

Bonus:

Label the cities 0, …, 10 starting from Rome and going in anticlockwise direction. Then, starting from city $i$, the number of clockwise steps ($c_i$) must be $i$ more than the number of anticlockwise steps ($a_i$) modulo 11, i.e., $c_i - a_i = i \mod 11$. But we also want the total number of steps to be the same starting from any city. In particular, $c_0 + a_0 = c_1 + a_1$. If $c_0 \ne a_0$, then we already need at least 11 steps, so we need $c_0 = a_0$. Then, $c_1 + a_1$ must be even, but then $c_1 - a_1 - 1$ must be an odd multiple of 11. If it is +11, then $c_1 + a_1 \ge 12$, which is more than the above solution. So it should be -11, in which case, $c_1+a_1 \ge 10$. For any other odd multiple of 11, the total number of steps is clearly larger. Therefore, the above solution is minimal.

This should work.

1. The colouring:

2. The sequence:

5 R's followed by 5 B's.

Bonus:

Label the cities 0, …, 10 starting from Rome and going in anticlockwise direction. Then, starting from city i, the number of clockwise steps (c_i) must be i more than the number of anticlockwise steps (a_i) modulo 11, i.e., c_i - a_i = i mod 11. But we also want the total number of steps to be the same starting from any city. In particular, c₀ + a₀ = c₁ + a₁. If c₀ ≠ a₀, then we already need at least 11 steps, so we must have c₀ = a₀. Then, c₁ + a₁ must be even, which means c₁ - a₁ = 11k+1, where k is odd. It follows that c₁ + a₁ ≥ |11k+1| ≥ 10 since k is odd. Therefore, the above solution is minimal.

This should work.

1. The colouring:

2. The sequence:

R, R, R, R, R, R, B, B, B, B, B, i.e., 6 R's followed by 5 B's.

Bonus:

Label the cities 0, …, 10 starting from Rome and going in anticlockwise direction. Then, starting from city $i$, the number of clockwise steps ($c_i$) must be $i$ more than the number of anticlockwise steps ($a_i$) modulo 11, i.e., $c_i - a_i = i \mod 11$. But we want the total number of steps to be the same starting from any city. In particular, $c_0 + a_0 = c_1 + a_1$. If $c_0 = a_0$, then $c_1 + a_1$ must be even, but then $c_1 - a_1 - 1$ must be an odd multiple of 11, which means $c_1 + a_1$ is at least 12 which is more than the above solution. Therefore, $c_0 - a_0$ is a nonzero multiple of 11 and the above solution is minimal.

This should work.

1. The colouring:

2. The sequence:

5 R's followed by 5 B's.

Bonus:

Label the cities 0, …, 10 starting from Rome and going in anticlockwise direction. Then, starting from city $i$, the number of clockwise steps ($c_i$) must be $i$ more than the number of anticlockwise steps ($a_i$) modulo 11, i.e., $c_i - a_i = i \mod 11$. But we also want the total number of steps to be the same starting from any city. In particular, $c_0 + a_0 = c_1 + a_1$. If $c_0 \ne a_0$, then we already need at least 11 steps, so we need $c_0 = a_0$. Then, $c_1 + a_1$ must be even, but then $c_1 - a_1 - 1$ must be an odd multiple of 11. If it is +11, then $c_1 + a_1 \ge 12$, which is more than the above solution. So it should be -11, in which case, $c_1+a_1 \ge 10$. For any other odd multiple of 11, the total number of steps is clearly larger. Therefore, the above solution is minimal.

This should work.

1. The colouring:

2. The sequence:

R, R, R, R, R, R, B, B, B, B, B, i.e., 6 R's followed by 5 B's.

This should work.

1. The colouring:

2. The sequence:

R, R, R, R, R, R, B, B, B, B, B, i.e., 6 R's followed by 5 B's.

Bonus:

Label the cities 0, …, 10 starting from Rome and going in anticlockwise direction. Then, starting from city $i$, the number of clockwise steps ($c_i$) must be $i$ more than the number of anticlockwise steps ($a_i$) modulo 11, i.e., $c_i - a_i = i \mod 11$. But we want the total number of steps to be the same starting from any city. In particular, $c_0 + a_0 = c_1 + a_1$. If $c_0 = a_0$, then $c_1 + a_1$ must be even, but then $c_1 - a_1 - 1$ must be an odd multiple of 11, which means $c_1 + a_1$ is at least 12 which is more than the above solution. Therefore, $c_0 - a_0$ is a nonzero multiple of 11 and the above solution is minimal.