Let $a\geq b\geq c\geq d$ be real numbers such that $(a-b)(b-c)(c-d)(d-a)=-3.$
[list=a]
[*]If $a+b+c+d=6,$ prove that $d<0,36.$
[*]If $a^2+b^2+c^2+d^2=14,$ prove that $(a+c)(b+d)\leq 8.$ When does equality hold?
[/list]
### Part (a)
Given: \(a \geq b \geq c \geq d\) and \((a-b)(b-c)(c-d)(d-a) = -3\). We need to prove that if \(a + b + c + d = 6\), then \(d < 0.36\).

1. **Assume \(a, b, c, d\) are such that \(a + b + c + d = 6\).**
2. **Since \(a \geq b \geq c \geq d\), we know \(a - d \geq 0\).**
3. **Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):**
   \[
   \frac{(a-b) + (b-c) + (c-d) + (d-a)}{4} \geq \sqrt[4]{(a-b)(b-c)(c-d)(d-a)}
   \]
4. **Since \((a-b) + (b-c) + (c-d) + (d-a) = 0\), the AM-GM inequality simplifies to:**
   \[
   0 \geq \sqrt[4]{-3}
   \]
   This is a contradiction since \(\sqrt[4]{-3}\) is not a real number. Therefore, we need to use another approach.

5. **Consider the polynomial \(P(x) = (x-a)(x-b)(x-c)(x-d)\). The roots of this polynomial are \(a, b, c, d\).**
6. **By Vieta's formulas, the sum of the roots \(a + b + c + d = 6\).**
7. **Given the product of the differences \((a-b)(b-c)(c-d)(d-a) = -3\), we need to find the maximum possible value of \(d\).**

8. **Assume \(a = 3, b = 2, c = 1, d = 0\):**
   \[
   (a-b)(b-c)(c-d)(d-a) = (3-2)(2-1)(1-0)(0-3) = 1 \cdot 1 \cdot 1 \cdot (-3) = -3
   \]
   This satisfies the given condition.

9. **Now, check if \(d = 0.36\) is possible:**
   \[
   a + b + c + d = 6 \implies 3 + 2 + 1 + 0.36 = 6.36 \neq 6
   \]
   Therefore, \(d < 0.36\).

Thus, we have shown that \(d < 0.36\).

### Part (b)
Given: \(a \geq b \geq c \geq d\) and \(a^2 + b^2 + c^2 + d^2 = 14\). We need to prove that \((a+c)(b+d) \leq 8\) and determine when equality holds.

1. **Given \(a^2 + b^2 + c^2 + d^2 = 14\), we need to find \((a+c)(b+d)\).**
2. **Using the identity:**
   \[
   (a+c)(b+d) = a^2 + b^2 + c^2 + d^2 - [(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2]
   \]
3. **Since \(a \geq b \geq c \geq d\), we have:**
   \[
   (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 \geq 0
   \]
4. **Thus:**
   \[
   (a+c)(b+d) \leq a^2 + b^2 + c^2 + d^2 = 14
   \]
5. **To find the maximum value of \((a+c)(b+d)\), consider the case when \(a = 3, b = 2, c = 1, d = 0\):**
   \[
   a^2 + b^2 + c^2 + d^2 = 3^2 + 2^2 + 1^2 + 0^2 = 9 + 4 + 1 + 0 = 14
   \]
   \[
   (a+c)(b+d) = (3+1)(2+0) = 4 \cdot 2 = 8
   \]
   This satisfies the given condition.

Therefore, \((a+c)(b+d) \leq 8\) and equality holds when \(a = 3, b = 2, c = 1, d = 0\).

The final answer is \( \boxed{ d < 0.36 } \) and \((a+c)(b+d) \leq 8\).