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1. Identify the second root using the conjugate root theorem.

Since the polynomial has real coefficients $p$ and $q$, the complex roots must occur in conjugate pairs.
Therefore, the second root is $1 - \sqrt{2}\mathrm{i}$.

2. Form the quadratic factor corresponding to the two complex roots.

The quadratic factor formed by the complex conjugate roots is:
$(x - (1 + \sqrt{2}\mathrm{i}))(x - (1 - \sqrt{2}\mathrm{i}))$
$= ((x - 1) - \sqrt{2}\mathrm{i})((x - 1) + \sqrt{2}\mathrm{i})$
$= (x - 1)^2 - (\sqrt{2}\mathrm{i})^2$
$= (x^2 - 2x + 1) - (-2)$
$= x^2 - 2x + 3$

3. Determine the third root by introducing a linear factor and comparing constant terms.

Let the remaining linear factor be $3x + c$. The cubic polynomial can be factored as:
$3x^3 + px^2 + qx + 3 = (x^2 - 2x + 3)(3x + c)$
Comparing the constant terms on both sides:
$3c = 3 \implies c = 1$
The linear factor is $3x + 1$. Setting this to $0$ gives the third root:
$3x + 1 = 0 \implies x = -\frac{1}{3}$

4. Find the values of $p$ and $q$ by expanding the polynomial factors.

Substitute $c = 1$ and expand the expression:
$(x^2 - 2x + 3)(3x + 1) = x^2(3x + 1) - 2x(3x + 1) + 3(3x + 1)$
$= 3x^3 + x^2 - 6x^2 - 2x + 9x + 3$
$= 3x^3 - 5x^2 + 7x + 3$
Comparing this with $3x^3 + px^2 + qx + 3$, we deduce the unknown coefficients:
$p = -5$
$q = 7$

1. Express $z^*$ in terms of $w$ and substitute into the first equation to form a quadratic equation.

From $w + z^* = 2 - \mathrm{i}$, we have:
$z^* = 2 - \mathrm{i} - w$

Substitute this into $wz^* = 3 - \mathrm{i}$:
$w(2 - \mathrm{i} - w) = 3 - \mathrm{i}$
$2w - \mathrm{i}w - w^2 = 3 - \mathrm{i}$
$w^2 - (2 - \mathrm{i})w + (3 - \mathrm{i}) = 0$

2. Solve the quadratic equation to find the possible values of $w$.

Using the quadratic formula for $w^2 - (2 - \mathrm{i})w + (3 - \mathrm{i}) = 0$:
$w = \frac{(2 - \mathrm{i}) \pm \sqrt{(2 - \mathrm{i})^2 - 4(1)(3 - \mathrm{i})}}{2}$
$w = \frac{2 - \mathrm{i} \pm \sqrt{4 - 4\mathrm{i} + \mathrm{i}^2 - 12 + 4\mathrm{i}}}{2}$
$w = \frac{2 - \mathrm{i} \pm \sqrt{4 - 4\mathrm{i} - 1 - 12 + 4\mathrm{i}}}{2}$
$w = \frac{2 - \mathrm{i} \pm \sqrt{-9}}{2}$
$w = \frac{2 - \mathrm{i} \pm 3\mathrm{i}}{2}$

This gives two possible values for $w$:
$w_1 = \frac{2 + 2\mathrm{i}}{2} = 1 + \mathrm{i}$
$w_2 = \frac{2 - 4\mathrm{i}}{2} = 1 - 2\mathrm{i}$

3. Find the corresponding values of $z$ for each possible value of $w$.

Case 1: If $w = 1 + \mathrm{i}$
$z^* = 2 - \mathrm{i} - (1 + \mathrm{i}) = 1 - 2\mathrm{i}$
Taking the conjugate yields $z = 1 + 2\mathrm{i}$.

Case 2: If $w = 1 - 2\mathrm{i}$
$z^* = 2 - \mathrm{i} - (1 - 2\mathrm{i}) = 1 + \mathrm{i}$
Taking the conjugate yields $z = 1 - \mathrm{i}$.

4. Calculate the modulus of $w$ and $z$ in both cases to apply the condition $|w| < |z|$ and determine the final valid pair.

Check Case 1: $w = 1 + \mathrm{i}, z = 1 + 2\mathrm{i}$
$|w| = \sqrt{1^2 + 1^2} = \sqrt{2}$
$|z| = \sqrt{1^2 + 2^2} = \sqrt{5}$
Since $\sqrt{2} < \sqrt{5}$, this pair satisfies $|w| < |z|$.

Check Case 2: $w = 1 - 2\mathrm{i}, z = 1 - \mathrm{i}$
$|w| = \sqrt{1^2 + (-2)^2} = \sqrt{5}$
$|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$
Since $\sqrt{5} \not< \sqrt{2}$, this pair is rejected.

Hence, $w = 1 + \mathrm{i}$ and $z = 1 + 2\mathrm{i}$.