In general, the rotational inertia of a rigid body is fully characterized by an inertia tensor, which has 6 independent variables: $I_{xx}$, $I_{yy}$, $I_{zz}$, $I_{xy}$, $I_{yz}$, $I_{zx}$. The three diagonal terms can be calculated from moments of inertia about the x,y,z axes (which are $I_x$, $I_y$, $I_z$ in your notation):
$I_{xx} = \frac{I_y + I_z - I_x}{2}$; $I_{yy} = \frac{I_z + I_x - I_y}{2}$; $I_{zz} = \frac{I_x + I_y - I_z}{2}$
Unfortunately, the three non-diagonal terms $I_{xy}$, $I_{yz}$, $I_{zx}$ cannot be calculated from $I_x$, $I_y$, $I_z$; they are independent quantities that must be measured separately. However, these terms are often zero when the object is symmetric: if there is reflective symmetry in the yz-plane, then $I_{xy} = I_{zx} = 0$; if there is reflective symmetry in the zx-plane, then $I_{yz} = I_{xy} = 0$. I suspect both of these are true for a human arm, to a good enough approximation for your purposes.
Once you know all the components of the inertia tensor, you can calculate the moment $I_n$ about an arbitrary unit vector $ \begin{pmatrix}
n_x\\
n_y\\
n_z
\end{pmatrix}$ using
$I_n = (I_{xx} + I_{yy} + I_{zz}) - I_{nn}$, where:
\begin{equation*}
I_{nn} =
\begin{pmatrix}
n_x\\
n_y\\
n_z
\end{pmatrix}^T
\begin{pmatrix}
I_{xx}&I_{xy}&I_{xz}\\
I_{yx}&I_{yy}&I_{yz}\\
I_{zx}&I_{zy}&I_{zz}
\end{pmatrix}
\begin{pmatrix}
n_x\\
n_y\\
n_z
\end{pmatrix}
\end{equation*}
Note: the tensor is always symmetric in the mathematical sense (even if the object has no geometric symmetry). This means the written order of the indices doesn't matter: $I_{xy} = I_{yx}$, etc. That is why the inertia tensor has 6 degrees of freedom, rather than 9.
(1) is definitely wrong. Moment of inertia of a body is $\int_{\forall dm}{}(dm) r^2$ and not $(\int_{\forall dm}^{} dm)r^2$ (you have to add the product of mass and distance, not add mass, then multiply with distance).
(2) is right but you can't find $I_{CM}$ just from the given information.
I really don't know what you did in (3). $I_{CM}$ is not $\frac{1}{2}MR^2$ in this case.
You cannot find the moment of inertia just by knowing the mass off the body and by how much the center of mass has been displaced i.e location of center of mass.
Let me give an example, lets say there are 2 masses of mass $m$. Now, the center of mass is at the midpoint of these two points. Lets say half of the distance between these two points is $r$. So, the moment of inertia of both the masses about the mid point is $2mr^2$. Now, move each of the masses by a distance $r$ further from the center of mass. The center of mass's position remains unchanged but the moment of inertia became $8mr^2$.
The point is, for a given location of center of mass and a mass $M$, there are infinitely many distributions of mass which give the same position of center of mass but different moment of inertia about the same point. You cannot deduce $\int_{\forall dm}{}(dm) r^2$ from $\int_{\forall dm}{}(dm) \vec{r}^{\,}$.
---------------------EDIT:--------------------
You did not mention before your edit that the moment of inertia about the center of mass was given.
You seem to be having a problem with the expression of kinetic energy of a rotating body. The total kinetic energy expression is $$K=K_{translational}+K_{roatational}=\frac{1}{2}Mv_{CM}^2+\frac{1}{2}I_{CM}\omega ^2$$
The roatational kinetic energy as such can be about any axis and varies from axis to axis because of change in moment of inertia. However, when you're calculating the total energy expression $K=K_{translational}+K_{roatational}=\frac{1}{2}Mv_{CM}^2+\frac{1}{2}I\omega ^2$ you have to use moment of inertia about $CM$. The reason becomes evident when you go through the derivation of the total kinetic energy of a rotating body. I'll give the derivation later if you want because that will be going out of scope of the question.
Also, mark you edit. Right now, it seems like its a continuation.
Best Answer
Strictly speaking, this isn't the parallel axis theorem, but simply the addition of three moments of inertia about a common axis.
$$\mathscr{I}_{total}=\mathscr{I}_{man/arms}+\mathscr{I}_{dumbell}+\mathscr{I}_{dumbell}$$
The parallel-axis theorem deals with a shift in rotational axis (theoretical or actual) from an axis through the center of mass. The axis never shifts in this problem except for possibly the idea that the dumbbells shift. But they are simply treated as point masses which have $\mathscr{I}=mr^2$, where $r$ is the distance from the dumbbell point mass to the axis.