Find the polynomials $ f(x)$ having the following properties:

(i) $ f(0) \equal{} 1$, $ f'(0) \equal{} f''(0) \equal{} \cdots \equal{} f^{(n)}(0) \equal{} 0$
(ii) $ f(1) \equal{} f'(1) \equal{} f''(1) \equal{} \cdots \equal{} f^{(m)}(1) \equal{} 0$
1. **Initial Conditions and Polynomial Form**:
   Given the conditions:
   \[
   f(0) = 1, \quad f'(0) = f''(0) = \cdots = f^{(n)}(0) = 0
   \]
   and
   \[
   f(1) = f'(1) = f''(1) = \cdots = f^{(m)}(1) = 0,
   \]
   we can infer that \( f(x) \) must be of the form:
   \[
   f(x) = x^n g(x) + 1 = (x-1)^m k(x).
   \]

2. **Equating the Two Forms**:
   Equating the two forms, we get:
   \[
   (x-1)^m k(x) - x^n g(x) = 1.
   \]
   This equation must hold for all \( x \).

3. **Greatest Common Divisor**:
   Since \(\gcd((x-1)^m, x^n) = 1\) for all \( m, n \), we can find polynomials \( K(x) \) and \( G(x) \) such that:
   \[
   (x-1)^m K(x) - x^n G(x) = 1.
   \]

4. **General Solution**:
   The general solution to the equation can be written as:
   \[
   k(x) = K(x) + h(x) x^n \quad \text{and} \quad g(x) = G(x) + h(x) (x-1)^m,
   \]
   for some polynomial \( h(x) \in \mathbb{R}[x] \).

5. **Form of \( f(x) \)**:
   Substituting back, we get:
   \[
   f(x) = x^n g(x) + 1 = x^n (G(x) + h(x) (x-1)^m) + 1.
   \]
   Simplifying, we get:
   \[
   f(x) = x^n G(x) + x^n h(x) (x-1)^m + 1.
   \]

6. **Canonical Form**:
   Therefore, the polynomials \( f(x) \) that satisfy the given conditions are of the form:
   \[
   f(x) = x^n G(x) + x^n h(x) (x-1)^m + 1,
   \]
   where \( G(x) \) and \( h(x) \) are polynomials in \( \mathbb{R}[x] \).

The final answer is \( \boxed{ f(x) = x^n G(x) + x^n h(x) (x-1)^m + 1 } \).