Calculating current in a parallel AC RL circuit
You attempt has many issues. First, let's clear up what \$\omega\$ is. With an expression in the form of \$ \sin(at) \$, one period or one cycle is when:
$$ at_{period} = 2\pi $$
$$\Rightarrow a = 2\pi\frac{1}{t_{period}} = 2\pi\times{freq} = \omega $$
Therefore, \$ \omega \$ is simply the coefficient in front of \$ t \$, which is \$10^3\$ in 4a).

For the circuit 4a):
$$ I_{total} = \frac{V}{Z_{total}} $$
$$ Z_{total} = \frac{1}{\frac{1}{Z_R}+\frac{1}{Z_L}} $$
(\$ Z_{total} \$ is not \$ Z_R + Z_L \$ as in your step 3.)

The equation is equivalent to:
$$ I_{total} = \frac{V}{Z_R} + \frac{V}{Z_L} = I_R + I_L $$
I will use this alternative representation because the intermediate quantities are slightly more interesting.
$$ I_R = \frac{100\angle 50^\circ}{5} = 20\angle 50^\circ $$
You are looking for an answer with time dependence, use the impedance of the inductor \$Z_L = j\omega L\$ which has the time related phase information (don't use reactance).
$$ I_L = \frac{100\angle 50^\circ}{(j\omega L = \omega L\angle 90^\circ)}
= \frac{100}{1000\times0.02} \angle(50-90)^\circ = 5\angle{-40}^\circ $$
Finally,
$$ I_{total} = 20\angle 50 + 5\angle {-40} = 20.6\angle 35.96 $$
(You need to look up how to add two numbers with phase angles. You cannot just add the amplitudes and the phase angles.)
$$ I_{total} = 20.6\sin(10^3t+35.96^\circ)\space A $$