The description presented looks confusing,
so here is one possible interpretation,
if the subject is about
making a $C_2$ connection
of the two cubic Bezier curves (segments).
The expression
\begin{align}
s_1(t)&=
B_0^3(t)\left[\begin{matrix}0\\ 0\end{matrix}\right]
+B_1^3(t)\left[\begin{matrix}0\\ 1\end{matrix}\right]
+B_2^3(t)\left[\begin{matrix}1\\ 2\end{matrix}\right]
+B_3^3(t)\left[\begin{matrix}2\\ 2\end{matrix}\right]
\tag{1}\label{1}
\end{align}
defines the coordinates of a point
$s_1(t)$
for any $t\in[0,1]$
on the first cubic Bezier curve (segment)
with control points
\begin{align}
p_{01}&=(0,0)
,\quad
p_{11}=(0,0)
,\quad
p_{21}=(1,2)
,\quad
p_{31}=(2,2)
,
\end{align}
where, for example, $p_{21}=(1,2)$
means that
$x$-coordinate of
of the point $p_{21}$ is $1$,
and its $y$-coordinate is $2$.
Points $p_{0k}$ and $p_{3k}$
are always located on the endpoints of the curve.
The other two points $p_{1k}$ and $p_{2k}$
usually are not on the curve,
but they control the shape of the curve.
Next, $B_k^3(t)$ means
$k$-th cubic
Bernstein polynomial,
\begin{align}
B_k^3(t)&=\textstyle\binom{3}{k}(1-t)^{3-k}t^k
,\\
B_0^3(t)&=(1-t)^3
,\quad
B_1^3(t)=3(1-t)^2t
,\quad
B_2^3(t)=3(1-t)t^2
,\quad
B_3^3(t)=t^3
.
\end{align}
Thus, the expression \eqref{1}
can be written in a general form
for the $k$-th segment as
\begin{align}
s_k(t)&=
(1-t)^3 p_{0k}
+3(1-t)^2t\cdot p_{1k}
+3(1-t)t^2\cdot p_{2k}
+t^3\cdot p_{3k}
\tag{2}\label{2}
,
\end{align}
which in fact represents two expressions (in 2D),
for $x$ and $y$-coordinates of the point $s_k(t)$
on the $k$-th Bezier segment
\begin{align}
s_{kx}(t)&=
(1-t)^3 p_{0kx}
+3(1-t)^2t\cdot p_{1kx}
+3(1-t)t^2\cdot p_{2kx}
+t^3\cdot p_{3kx}
,\\
s_{ky}(t)&=
(1-t)^3 p_{0ky}
+3(1-t)^2t\cdot p_{1ky}
+3(1-t)t^2\cdot p_{2ky}
+t^3\cdot p_{3ky}
,
\end{align}
which are just ordinary cubic polynomials in terms of parameter $t$.
Note that for any segment the parameter
$t$ (often called as time parameter) runs from
$t=0$ (point $p_{0k}$)
to
$t=1$ (point $p_{3k}$).
The sequence of such connected segments $s_k$ is often called
a cubic Bezier spline
(don't confuse it with B-splines!),
and the word "spline" suggests that these segments
are connected smoothly,
which this exercise was supposed to demonstrate.
What is missing in the original illustration,
are the two actual cubic Bezier segments,
which are smoothly connected:
So, the instructions tell us that
given the first cubic Bezier segment $s_1$
which is defined by control points
$p_{01}$ through $p_{31}$,
we can smoothly
(that is, the resulting two-segment curve will be $C_2$-continuous)
extend the curve with segment $s_2$,
by calculating three of its control points,
$p_{02},p_{12}$ and $p_{22}$.
The last control point can be chosen freely,
its location does not affect
the smoothness of the curve $s_1,s_2$
(in fact, the point $p_{01}$ can be chosen freely as well).
In order to get the $C_2$-continuity, we need first ensure
$C_0$, that is, the starting point
$p_{02}$ of the second segment
must coincide with the endpoint $p_{31}$ of the first:
\begin{align}
p_{02}&=p_{31}
,\\
\text{or }\quad
s_1(1)&=s_2(0)
.
\end{align}
To get $C_1$-continuity, we need
to guarantee that
\begin{align}
s'_1(t)|_{t=1}&=s'_2(t)|_{t=0}
,\\
\text{which gives }\quad
p_{12} &= 2p_{31}-p_{21}
\end{align}
The last condition for $C_2$--continuity is
\begin{align}
s''_1(t)|_{t=1}&=s''_2(t)|_{t=0}
,\\
\text{which gives }\quad
p_{22} &= p_{11}+2p_{12}-2p_{21}
.
\end{align}
Also, note that the phrase
"$\dots$ which connects to point $(2,2)$
with a $C_2$-continuity, over $[1, 2]$"
is incorrect.
The second segment (the actual curve)
smoothly connects the endpoint $p_{31}$
of the first segment $s_1$
with the endpoint $p_{32}$
of the second segment $s_2$,
and not "over $[1, 2]$",
but still "over $[0, 1]$",
that is, $t\in[0,1]$.
Even if, for example, this two-segment curve
represents a smooth trajectory of the point
for some timeline $T$ and
we assume that
the segment $s_1$ starts at $T=0$
and ends at $T=1$,
then $s_2$ follows and ends at $T=2$,
we still have to use $t$ from 0 to 1
to calculate the points
according to \eqref{2}.
Edit
In case if we would like to use
a global time parameter $T$
to control the smooth movement of the point
using given sequence of "time points",
we need to use modified expression
for the point on the curve.
Since cubic Bezier curves are defined
for local time parameter $t\in[0,1]$,
we need to modify the procedure
of the smooth connection of the two segments
by using a transformation
from the global $T$,
starting time $T=t_k$
and ending time $T=t_{k+1}$
for the segment $s_k$:
\begin{align}
t&=\tfrac{T-t_k}{t_{k+1}-t_k}
.
\end{align}
Then we have the expression
for the coordinated of the point
on the segment $s_k$ as
\begin{align}
s_k(T)
&=
(1-\tfrac{T-t_k}{t_{k+1}-t_k})^3\cdot p_{0k}
\\
&+3\cdot(1-\tfrac{T-t_k}{t_{k+1}-t_k})^2(\tfrac{T-t_k}{t_{k+1}-t_k})\cdot p_{1k}
\\
&+3\cdot(1-\tfrac{T-t_k}{t_{k+1}-t_k})(\tfrac{T-t_k}{t_{k+1}-t_k})^2\cdot p_{2k}
\\
&+(\tfrac{T-t_k}{t_{k+1}-t_k})^3\cdot p_{3k}
.
\end{align}
Given that, we can find the control points
of the smoothly attached segment $s_{k+1}$,
which has assigned global time length $t_{k+2}-t_{k+1}$,
that satisfy three conditions of continuity:
\begin{align}
s_{k+1}(T)|_{T=t_{k+1}}&=s_{k}(T)|_{T=t_{k+1}}
,\\
s'_{k+1}(T)|_{T=t_{k+1}}&=s'_{k}(T)|_{T=t_{k+1}}
,\\
s''_{k+1}(T)|_{T=t_{k+1}}&=s''_{k}(T)|_{T=t_{k+1}}
.
\end{align}
This system results in
\begin{align}
p_{0\,k+1}&=p_{1\,k}
,\\
p_{1\,k+1}&=
p_{3\,k}-
\tfrac{t_{k+2}-t_{k+1}}{t_{k+1}-t_k}\cdot(p_{2\,k}-p_{3\,k})
,\\
p_{2\,k+1}&=
2p_{1\,k+1}-p_{3\,k}
+(\tfrac{t_{k+2}-t_{k+1}}{t_{k+1}-t_k})^2\cdot(p_{1\,k}-2\,p_{2\,k}+p_{3\,k})
.
\end{align}
Back to the OP: assuming we have
\begin{align}
t_0&=0
,\quad
t_{1}=1
,\quad
t_{2}=3
.
\end{align}
Then the control points of the extension curve $q_2$ would be
\begin{align}
q_{02}&=p_{31}
,\\
q_{12}&=3\,p_{31}-2\,p_{21}
,\\
q_{22}&=4\,p_{11}-12\,p_{21}+9\,p_{31}
.
\end{align}
And since we still can choose any point as $q_{32}$,
let $q_{32}=p_{32}$, so we can compare
the two extension segments, which differ
in the time span: for the first one we have
$t_1=1$, $t_2=2$, and
for the second one we have
$t_1=1$, $t_2=3$.
Here is the difference:
