Step 1: List the sample space.
Two dice give \(6 \times 6 = 36\) equally likely outcomes.
We need outcomes where both numbers are even AND their sum is 5 or more.
Step 2: List the even-even outcomes.
Each die shows 2, 4 or 6, giving \(3 \times 3 = 9\) outcomes: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6).
Step 3: Apply the sum condition.
The pair (2,2) sums to 4, which fails the "sum 5 and above" condition.
All the other 8 pairs, (2,4)=6, (2,6)=8, (4,2)=6, (4,4)=8, (4,6)=10, (6,2)=8, (6,4)=10, (6,6)=12, satisfy sum \(\ge 5\).
So 8 out of the 36 total outcomes meet both conditions.
Final Answer:
The probability is \(\dfrac{8}{36} = 0.2\overline{2}\), rounded to 0.22.
\[ \boxed{0.22} \]