Given an unfair 6 sided dice. When the dice is rolled up two times, let $ P$ be the probability such that same number comes out and $ Q$ be the one such that the number of odd in the first and that of even in the second.

(1) Prove that $ P\geq \frac{1}{6}$,  then find the necessary and sufficient condition for which the equality holds.

(2) Prove that $ \frac{1}{4}\geq Q\geq \frac{1}{2}\minus{}\frac{3}{2}P$.
Given an unfair 6-sided die, we need to prove certain inequalities involving probabilities \( P \) and \( Q \).

1. **Prove that \( P \geq \frac{1}{6} \) and find the necessary and sufficient condition for which the equality holds.**

   Let \( a, b, c, d, e, f \) be the probabilities of rolling 1, 2, 3, 4, 5, and 6 respectively. These probabilities satisfy:
   \[
   a + b + c + d + e + f = 1
   \]
   The probability \( P \) that the same number comes out when the die is rolled twice is:
   \[
   P = a^2 + b^2 + c^2 + d^2 + e^2 + f^2
   \]

   By the Cauchy-Schwarz inequality in the form:
   \[
   (a^2 + b^2 + c^2 + d^2 + e^2 + f^2)(1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2) \geq (a + b + c + d + e + f)^2
   \]
   Simplifying, we get:
   \[
   6(a^2 + b^2 + c^2 + d^2 + e^2 + f^2) \geq 1^2
   \]
   \[
   a^2 + b^2 + c^2 + d^2 + e^2 + f^2 \geq \frac{1}{6}
   \]
   Therefore, \( P \geq \frac{1}{6} \).

   The equality holds if and only if \( a = b = c = d = e = f = \frac{1}{6} \). This is the necessary and sufficient condition for equality.

2. **Prove that \( \frac{1}{4} \geq Q \geq \frac{1}{2} - \frac{3}{2}P \).**

   The probability \( Q \) that the first roll is odd and the second roll is even is given by:
   \[
   Q = (a + c + e)(b + d + f)
   \]

   **Left Inequality: \( Q \leq \frac{1}{4} \)**

   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
   \[
   \frac{(a + c + e) + (b + d + f)}{2} \geq \sqrt{(a + c + e)(b + d + f)}
   \]
   Since \( a + b + c + d + e + f = 1 \), we have:
   \[
   \frac{1}{2} \geq \sqrt{(a + c + e)(b + d + f)}
   \]
   Squaring both sides:
   \[
   \frac{1}{4} \geq (a + c + e)(b + d + f)
   \]
   Therefore, \( Q \leq \frac{1}{4} \).

   **Right Inequality: \( Q \geq \frac{1}{2} - \frac{3}{2}P \)**

   Applying the Cauchy-Schwarz inequality to the sum of squares:
   \[
   (a + d)^2 + (a + e)^2 + (a + f)^2 + (b + d)^2 + (b + e)^2 + (b + f)^2 + (c + d)^2 + (c + e)^2 + (c + f)^2 \geq (a + b + c + d + e + f)^2
   \]
   Simplifying, we get:
   \[
   3(a^2 + b^2 + c^2 + d^2 + e^2 + f^2) + 2(a + b + c)(d + e + f) \geq 1
   \]
   \[
   3P + 2Q \geq 1
   \]
   \[
   2Q \geq 1 - 3P
   \]
   \[
   Q \geq \frac{1}{2} - \frac{3}{2}P
   \]

The final answer is \( \boxed{ P \geq \frac{1}{6} } \) and \( \frac{1}{4} \geq Q \geq \frac{1}{2} - \frac{3}{2}P \).