The above posts are certainly sufficient answers, but for fun I thought I would precisely characterize the set of points that can be reached from the origin.
Note that performing a move of type (1) amounts to adding $(2,1)$ to the particle's position vector, and, likewise, a move of type (2) amounts to adding $(1,2)$ to the particle's position vector. Therefore the points the particle can reach from the origin are precisely those points of the form
$$
n(1,2) + m(2,1) = (n + 2m, 2n + m)
$$
for nonnegative integers $n,m$. This is a precise characterization of the accessible points, but it isn't very easy to check.
It turns out that the much easier to check set of conditions
$a + b$ is divisible by $3$
$2a \geq b \geq \frac{a}{2}$
are necessary and sufficient for a point $(a,b)$ to be accessible.
To see this, suppose $(a,b)$ is accessible. Then there exist nonnegative integers $n,m$ such that
\begin{align}
n + 2m &= a\\
2n + m &= b
\end{align}
Thus
$$
a + b = 3n + 3m = 3(n + m)
$$
which, since $n,m$ are nonnegative integers, shows that $a + b$ is divisible by $3$. Solving for $n$ and $m$, we find
\begin{align}
n &= \frac{1}{3}(2b - a) \\
m &= \frac{1}{3}(2a - b)
\end{align}
Since $n,m$ are nonnegative integers, we have
\begin{align}
0 &\leq \frac{1}{3}(2b - a) \\
0 &\leq \frac{1}{3}(2a - b)
\end{align}
The first of these inequalities implies $b \geq \frac{a}{2}$, and the second implies $b \leq 2a$, i.e. $2a \geq b \geq \frac{a}{2}$.
Conversely, suppose that conditions 1 and 2 hold for some point $(a,b)$. By condition 1, there exists an integer $k$ such that $a + b = 3k$. Take $n = b - k$ and $m = a - k$. Note that $k = \frac{a + b}{3}$. We compute
\begin{align}
n &= b - k = b - \frac{a + b}{3} = \frac{2b - a}{3} \geq \frac{2\frac{a}{2} - a}{3} = \frac{a - a}{3} = 0 \\
m &= a - k = a - \frac{a + b}{3} = \frac{2a - b}{3} \geq \frac{2a - 2a}{3} = 0
\end{align}
Thus, $n$ and $m$ are nonnegative integers. Moreover
\begin{align}
(n + 2m, 2n + m) &= ((b - k) + 2(a - k) , 2(b - k) + (a - k)) = (b + 2a - 3k, 2b + a - 3k)
\\&= (b + a + a - 3k, b + b + a - 3k) = (3k + a - 3k, b + 3k - 3k) = (a,b)
\end{align}
By our original characterization of the accessible points, this shows that $(a,b)$ is an accessible point.
