Average alternating power
Using phasor (Steinmetz') notation with peak values (not RMS):

\$ V = 250 \angle{45°}\$
\$ I = 4 \angle{-30°}\$

Complex power is \$ S=\dfrac{1}{2}\; V\cdot I^{*}\$ where the asterisk means "complex conjugate", an operation which inverts the sign of the phase. Therefore:

\$ S = \dfrac{1}{2}\; (250 \angle{45°}) \cdot (4 \angle{-30°})^{*} = \dfrac{1}{2}\; (250 \angle{45°}) \cdot (4 \angle{30°}) = 500\angle{75°}\$

Analogously for the second example:

\$ S = \dfrac{1}{2}\; (18 \angle{-30°}) \cdot (5 \angle{-75°})^{*} = \dfrac{1}{2}\; (18 \angle{-30°}) \cdot (5\angle{75°}) = 45\angle{45°}\$

So effectively it seems there is an incoherence somewhere. Maybe a typo? In the second case if V had a phase of +30° the results would match.

EDIT

Just to incorporate and expand a bit of theory I explained in a comment to the OP.

Let's consider a linear load driven by a sinusoidal current \$i(t)\$ and having the voltage \$v(t)\$ across its terminals; because of the linearity of the load the voltage is sinusoidal too:

\begin{align*}
  v(t) &amp;= V_m \; \cos(\omega t + \phi_V) 
   &amp;&amp; \Leftrightarrow 
   &amp; V &amp;= V_m \, \angle \phi_V = V_m \, e^{j\phi_V}  \\
  i(t) &amp;= I_m \; \cos(\omega t + \phi_I) 
   &amp;&amp; \Leftrightarrow 
   &amp; I &amp;= I_m \, \angle \phi_I = I_m \, e^{j \phi_I}  \\
\end{align*}

Assuming the directions of i and v are associated, the instantaneous power absorbed by the load is:

\begin{align*}
  p(t) &amp;= v(t) \cdot i(t) 
  = V_m \; \cos(\omega t + \phi_V) \cdot I_m \; \cos(\omega t + \phi_I) = \\[1em]
  &amp;= \dfrac{1}{2} \, V_m \, I_m\; 
   \left[{ 
   \cos(2\omega t + \phi_V + \phi_I) + \cos(\phi_V - \phi_I)
   }\right]
\end{align*}

where the trig formula 
\$ \quad \cos(a)\cos(b) = \dfrac{1}{2}[cos(a+b) + cos(a-b)] \quad\$ 
was used.

The average power \$P\$, since \$p(t)\$ is periodic, can be computed averaging on a single period, thus:

\begin{align*}
  P &amp;= \dfrac{1}{T} \int_0^T {p(t)}\, dt 
  = \dfrac{1}{T} \int_0^T \dfrac{1}{2} \, V_m \, I_m\; 
   \left[{ 
   \cos(2\omega t + \phi_V + \phi_I) + \cos(\phi_V - \phi_I)
   }\right]\, dt = \\[1em]
  &amp;= \dfrac{1}{T} \int_0^T \dfrac{V_m \, I_m}{2} \cos(2\omega t + \phi_V + \phi_I) \, dt 
   + \dfrac{1}{T} \int_0^T \dfrac{V_m \, I_m}{2} \cos(\phi_V - \phi_I) \, dt
\end{align*}

The first integral is zero, because it is the integral of a sinusoidal function over its period, while the second integral is the average of a constant, so it is that constant. Therefore the average power becomes:

\begin{align*}
  P = \dfrac{V_m \, I_m}{2} \cos(\phi_V - \phi_I) = \dfrac{V_m \, I_m}{2} \cos \phi
\end{align*}

where \$ \phi = \phi_V - \phi_I \$ is the phase difference between the voltage and the current.

Without entering in further details, the average power is also called active power, whereas the reactive power is defined as:

\begin{align*}
  Q = \dfrac{V_m \, I_m}{2} \sin \phi
\end{align*}

On the other hand, if we define the complex power S as :

\begin{align*}
  S = \dfrac{1}{2} \, V \cdot I^{*}
\end{align*}

we obtain:

\begin{align*}
  S &amp;= \dfrac{1}{2} \, V_m \, e^{j \phi_V} \cdot (I_m \, e^{j \phi_I})^{*}
  = \dfrac{1}{2} \, V_m \, e^{j \phi_V} \cdot I_m \, e^{-j \phi_I}
  = \dfrac{V_m \, I_m}{2} \, e^{j (\phi_V - \phi_I)}
  = \dfrac{V_m \, I_m}{2} \, e^{j \phi} = \\[1em]
  &amp;= \dfrac{V_m \, I_m}{2} \cos \phi + j \, \dfrac{V_m \, I_m}{2} \sin \phi
  = P + j \, Q
\end{align*}

Hence it's easy to see why P and Q are respectively the real and the imaginary part of S.