Let $m \neq -1$ be a real number. Consider the quadratic equation 
$$
(m + 1)x^2 + 4mx + m - 3 =0.
$$
Which of the following must be true?

$\quad\rm(I)$ Both roots of this equation must be real.

$\quad\rm(II)$ If both roots are real, then one of the roots must be less than $-1.$

$\quad\rm(III)$ If both roots are real, then one of the roots must be larger than $1.$

$$
\mathrm a. ~ \text{Only} ~(\mathrm I)\rm \qquad \mathrm b. ~(I)~and~(II)\qquad \mathrm c. ~Only~(III) \qquad \mathrm d. ~Both~(I)~and~(III) \qquad \mathrm e. ~(I), (II),~and~(III)
$$
1. **Determine the discriminant to check if both roots are real:**

   The quadratic equation is given by:
   \[
   (m + 1)x^2 + 4mx + m - 3 = 0
   \]
   The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   Here, \(a = m + 1\), \(b = 4m\), and \(c = m - 3\). Substituting these values into the discriminant formula, we get:
   \[
   \Delta = (4m)^2 - 4(m + 1)(m - 3)
   \]
   Simplifying the expression:
   \[
   \Delta = 16m^2 - 4[(m + 1)(m - 3)]
   \]
   \[
   \Delta = 16m^2 - 4[m^2 - 3m + m - 3]
   \]
   \[
   \Delta = 16m^2 - 4[m^2 - 2m - 3]
   \]
   \[
   \Delta = 16m^2 - 4m^2 + 8m + 12
   \]
   \[
   \Delta = 12m^2 + 8m + 12
   \]
   Since \(12m^2 + 8m + 12\) is a quadratic expression in \(m\), we need to check if it is always positive. The discriminant of this quadratic expression is:
   \[
   \Delta' = 8^2 - 4 \cdot 12 \cdot 12 = 64 - 576 = -512
   \]
   Since \(\Delta' < 0\), the quadratic expression \(12m^2 + 8m + 12\) has no real roots and is always positive. Therefore, \(\Delta > 0\) for all \(m\), implying that the quadratic equation always has two distinct real roots.

2. **Check if one of the roots is less than \(-1\):**

   To determine the nature of the roots, we use the fact that the sum and product of the roots of the quadratic equation \(ax^2 + bx + c = 0\) are given by:
   \[
   \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a}
   \]
   For our equation, \(a = m + 1\), \(b = 4m\), and \(c = m - 3\). Thus:
   \[
   \alpha + \beta = -\frac{4m}{m + 1}
   \]
   \[
   \alpha \beta = \frac{m - 3}{m + 1}
   \]
   We need to check if one of the roots is less than \(-1\). Consider the value of the polynomial at \(x = -1\):
   \[
   P(-1) = (m + 1)(-1)^2 + 4m(-1) + (m - 3)
   \]
   \[
   P(-1) = m + 1 - 4m + m - 3
   \]
   \[
   P(-1) = -2m - 2
   \]
   Since \(P(-1) = -2(m + 1)\), and given \(m \neq -1\), \(P(-1)\) has the opposite sign to the leading coefficient \(m + 1\). This implies that \(-1\) is between the two roots, meaning one root is less than \(-1\) and the other is greater than \(-1\).

3. **Check if one of the roots is greater than \(1\):**

   To check if one of the roots is greater than \(1\), consider the value of the polynomial at \(x = 1\):
   \[
   P(1) = (m + 1)(1)^2 + 4m(1) + (m - 3)
   \]
   \[
   P(1) = m + 1 + 4m + m - 3
   \]
   \[
   P(1) = 6m - 2
   \]
   For \(P(1) > 0\), we need:
   \[
   6m - 2 > 0
   \]
   \[
   6m > 2
   \]
   \[
   m > \frac{1}{3}
   \]
   Therefore, if \(m > \frac{1}{3}\), one of the roots is greater than \(1\). However, this is not always true for all \(m\). For example, when \(m = 3\), the roots are \(\{-3, 0\}\), neither of which is greater than \(1\).

Conclusion:
- Statement (I) is true because the discriminant is always positive, ensuring two distinct real roots.
- Statement (II) is true because one root is always less than \(-1\).
- Statement (III) is not necessarily true as shown by the counterexample when \(m = 3\).

The final answer is \boxed{(I) \text{ and } (II)}.