To find the minimum possible value of \( U \):

1. Given the interquartile range (IQR) is 20 grams:

   \[ IQR = U – L = 20 \]

2. For no outliers, the maximum value must satisfy:

   \[ \text{Maximum} \leq U + 1.5 \times IQR \]

3. From the box plot, maximum value is 75 grams:

   \[ 75 \leq U + 1.5 \times 20 \]
   \[ 75 \leq U + 30 \]
   \[ U \geq 45 \]

Thus, the minimum possible value of \( U \) is \( \boxed{45} \) grams.

Using the result from part (a) and the given IQR:

1. From \( IQR = U – L = 20 \):

   \[ L = U – 20 \]

2. Using the minimum \( U = 45 \):

   \[ L = 45 – 20 = 25 \]

3. Verify no lower outliers:

   \[ \text{Minimum} \geq L – 1.5 \times IQR \]
   \[ \text{Minimum} \geq 25 – 30 = -5 \] (which is automatically satisfied as weights can’t be negative)

Thus, the minimum possible value of \( L \) is \( \boxed{25} \) grams.