Sum of probabilities must equal 1:

\( p + p + p + \frac{1}{2}p = 1 \)
\( \frac{7}{2}p = 1 \)
\( p = \boxed{\frac{2}{7}} \)

Expected value calculation:

\[ E(X) = 1×\frac{2}{7} + 2×\frac{2}{7} + 3×\frac{2}{7} + 4×\frac{1}{7} \]
\[ = \frac{2}{7} + \frac{4}{7} + \frac{6}{7} + \frac{4}{7} = \frac{16}{7} \]
\[ E(X) = \boxed{\frac{16}{7}} \]

(i) Since r is a probability: \[ \boxed{0 ≤ r ≤ 1} \]

(ii) Sum of probabilities for die B: \[ 3q + r = 1 ⇒ r = 1-3q \]
Since \( 0 ≤ r ≤ 1 \):
\[ 0 ≤ 1-3q ≤ 1 \]
\[ \boxed{0 ≤ q ≤ \frac{1}{3}} \]

\[ E(Y) = 1×q + 2×q + 3×q + 4×r = 6q + 4r \]
Using \( r = 1-3q \):
\[ E(Y) = 6q + 4(1-3q) = 4 – 6q \]
When \( q = 0 \): \( E(Y) = 4 \)
When \( q = \frac{1}{3} \): \( E(Y) = 2 \)
Thus: \[ \boxed{2 ≤ E(Y) ≤ 4} \]

Probability Agnes < Barbara is ½. There are 6 possible outcome pairs where A < B:

\[ P(A<B) = \frac{6}{7}q + \frac{6}{7}r = \frac{1}{2} \]
Substitute \( r = 1-3q \):
\[ \frac{6}{7}q + \frac{6}{7}(1-3q) = \frac{1}{2} \]
\[ \frac{6}{7} – \frac{12}{7}q = \frac{1}{2} \]
Solving gives \( q = \frac{5}{24} \), \( r = \frac{3}{8} \)

Now calculate E(Y):
\[ E(Y) = 4 – 6q = 4 – 6×\frac{5}{24} = \frac{11}{4} \]
\[ \boxed{E(Y) = \frac{11}{4}} \]