Determining the inital phase of a waveform
B leads C (or C lags B) because B crosses 0 on the same slope before C or B reaches a positive peak before C, etc.
To determine the phase angle with a scope, use the formula:
$$ \frac {d} {D} \times 360°$$
B has a period of 4 divisions or 360°. D = 4 divisions.  You could also use actual time, but the formula is a ratio.

Little d is where the second waveform crosses the x-axis on the same slope.  In this case d = 1.5 divisions.
$$ \frac {d} {D} \times 360° = \frac {1.5\ divisions} {4\ divisions} \times 360° = 135° $$
Phase Angle is 135°, which means B leads C by 135°.  At the same time C lags B by 135°.  Think of it like a queue.  First person always leads the second person.  2nd person is always behind the first.
Average power is:
$$ P = p(t) = v(t)\ i(t)$$
Normalized power assumes a load resistance of \$1\Omega \$.  This means current is numerically equal to voltage, so normalized power is:
$$ P_N = v(t)^2$$
$$ V_{RMS} = \sqrt {P_N}$$
$$ V_{MAX} = \sqrt {P_N}\ \sqrt {2}= \sqrt {2\ P_N}$$
which applies to any sinusoid (assuming a load resistance of \$1\Omega \$).