For two varieties $X,Y$ in projective space, we define $J(X,Y)$ the join of $X,Y$, to be the union of all lines in $\Bbb P^n$ connecting distinct points in $X$ and $Y$. Now I claim that $\pi(X) = J(X,P)\cap H$, because both sides represent taking the lines through $X$ and $P$ and then intersecting them with $H$. So by your work in (3), it suffices to determine $\dim J(X,P)\cap H$.
We can get rid of the intersection with $H$ in the dimension calculation easily, via the projective dimension theorem:
Projective dimension theorem (ref Hartshorne I.7.2): Let $X,Y$ be two irreducible closed subvarieties of $\Bbb P^n$ of codimensions $r,s$ respectively. Then every irreducible component of $X\cap Y$ has codimension at most $r+s$, and if $r+s\leq n$ then this intersection is nonempty.
If we know that $J(X,P)$ is irreducible, then as $P\notin H$, we see that $J(X,P)\cap H$ is a proper closed subvariety of $J(X,P)$, so it must have dimension at most $\dim J(X,P)-1$. On the other hand, by the theorem, it has dimension at least $\dim J(X,P)-1$. So we get $\dim J(X,P)\cap H = \dim J(X,P)-1$.
Now all we need to do is to prove that $J(X,P)$ is irreducible and determine it's dimension. Here we get a little bit of casework: in case (3), the join variety is just $X$ again, so it's irreducible of dimension $\dim X$. In cases (1) and (2), the following applies. Let $$J'(X,Y)=\{(x,y,z)\subset X\times Y\times \Bbb P^n \mid x\neq y, z\in [x,y]\}$$ where $[x,y]$ denotes the line passing through $x$ and $y$. Then $J(X,Y)$ is the projection of the closure of $J'(X,Y)$ to the final factor of $\Bbb P^n$. On the other hand, we can consider the projection of $\overline{J'(X,Y)}$ to $X\times Y$. The fibers of this projection are lines, thus irreducible of dimension 1. As a closed map with irreducible target and irreducible fibers must have irreducible source, we see that $J(X,Y)$ is irreducible and of dimension $\dim X + \dim Y + 1$. In our case, $Y$ is a point which has dimension zero, so $\dim J(X,P) = \dim X + 1$.
Subtracting 1 via the projective dimension theorem, we get the desired result in each case.
Best Answer
Yes, this is true, and you may find a suitable linear $\Bbb P^{n-r-1}$ by repeatedly projecting.
Ingredients:
Recipe:
Project $X$ from $p$ to $\Bbb P^{n-1}$. If the image of $X$ under this projection is $\Bbb P^{n-1}$, then $X$ was of dimension $n-1$ and the point $p$ suffices. If not, we can find another point $p_1$ in $\Bbb P^{n-1}$ not in the projection of $X$ and a $\Bbb P^{n-2}\subset \Bbb P^{n-1}$ not containing $p_1$, and project from $p$ to $\Bbb P^{n-2}$.
If the image of $X$ under this projection is $\Bbb P^{n-2}$, then $X$ was of dimension $n-2$ and the preimage of $p_1$ under the projection to $\Bbb P^{n-1}$ is a line in $\Bbb P^n$ not intersecting $X$, which works. If not, we can find another point $p_2$ in $\Bbb P^{n-2}$ not in the projection of $X$, and repeat the process.
Eventually, when we're projecting on to a linear subspace of dimension $\dim X$ we'll find that we get that the image of $X$ under this composite projection is all of the $\Bbb P^{\dim X}$ and the preimage in $\Bbb P^n$ of the last point we projected from is a linear subspace of dimension $n-r-1$ which doesn't intersect $X$.