You have an unbiased six-sided die and an unbiased three-sided die, so are drawing pairs uniformly distributed on $\{1,2,3,4,5,6\}\times\{0,1,2\}$.
$$\mathsf P(X+Y=1) ~=~ \mathsf P((X,Y)\in\{(1,0)\qquad\;\;\,\qquad\;\;\,\})~=~ 1/18
\\\mathsf P(X+Y=2) ~=~ \mathsf P((X,Y)\in\{(2,0),(1,1)\qquad\;\;\,\})~=~ 2/18
\\\mathsf P(X+Y=3) ~=~ \mathsf P((X,Y)\in\{(3,0),(2,1),(1,2)\})~=~ 3/18
\\\mathsf P(X+Y=4) ~=~ \mathsf P((X,Y)\in\{(4,0),(3,1),(2,2)\})~=~ 3/18
\\\mathsf P(X+Y=5) ~=~ \mathsf P((X,Y)\in\{(5,0),(4,1),(3,2)\})~=~ 3/18
\\\mathsf P(X+Y=6) ~=~ \mathsf P((X,Y)\in\{(6,0),(5,1),(4,2)\})~=~ 3/18
\\\mathsf P(X+Y=7) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,(6,1),(5,2)\})~=~ 2/18
\\\mathsf P(X+Y=8) ~=~ \mathsf P((X,Y)\in\{\qquad\;\;\,\qquad\;\;\,(6,2)\})~=~ 1/18
\\$$
An approach to this problem, a bit lengthy but having the advantage to provide a clear picture,
might be the following.
Start from considering the dice marked.
The set of possible, equi-probable, outcomes is represented by $6^3=216$ triples.
Then consider that you want the outcome of die 2 to be consecutive to the outcome of die 1, while the outcome of die 3 can be whatever
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 5} & {k + 1} & \forall \\
{prob} & & {5/6} & {1/6} & 1 \\
\end{array}
$$
the probability of getting such a scheme is $5/36$.
Now, since in our problem order does not matter, we shall swap (permute) the above configuration.
But we cannot perform that without considering the value of die 3 (call it $j$) compared with the result of die 1 and 2.
In fact, while e.g. $(1,2,3)$ can be permuted in $6$ ways, $(1,1,3)$ can be permuted in $3$ ways.
Moreover, we shall exclude the permuted triples that fall within the range of those already considered.
So, the prospect of the possible ordered configurations and number of ways to swap them is the following
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k + 1,\;k + 1 < j} \right\}} & {4 + 3 + 2 + 1} & {3!} & {60} \\
{\left\{ {k,\;k + 1,\;k + 1 = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;k = j} \right\}} & 5 & 3 & {15} \\
{\left\{ {k,\;k + 1,\;j = k - 1} \right\}} & 4 & 0 & 0 \\
{\left\{ {k,\;k + 1,\;j < k - 1} \right\}} & {1 + 2 + 3} & {3!} & {36} \\
\hline
{{\rm at}\,{\rm least}\,{\rm 2}\,{\rm consecutive}} & {} & {\rm = } & {126} \\
\end{array}
} $$
We see that the fourth configuration is cancelled as being already included in the first.
The prospect for the complementary case (no consecutive outcomes) will give
$$ \bbox[lightyellow] {
\begin{array}{*{20}c}
{config.} & {N.\,ordered} & {N.\,swaps} & {Tot.} \\
\hline
{\left\{ {k,\;k,\;k} \right\}} & 6 & 1 & 6 \\
{\left\{ {k,\;k,\; \ge k + 2} \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\; \ge k + 2,\; = } \right\}} & {4 + 3 + 2 + 1} & 3 & {30} \\
{\left\{ {k,\;k + 2,\; \ge k + 4} \right\}} & {2 + 1} & {3!} & {18} \\
{\left\{ {k,\;k + 3,\; \ge k + 5} \right\}} & 1 & {3!} & 6 \\
\hline
{{\rm none}\,{\rm consecutive}} & {} & {\rm = } & {90} \\
\end{array}
} $$
In the case of asking for three consecutive outcomes instead, considering them to be ordered we will have
$$
\begin{array}{c|ccc}
{die} & & 1 & 2 & 3 \\
{result} & & {1 \le k \le 4} & {k + 1} & {k + 2} \\
{prob} & & {4/6} & {1/6} & {1/6} \\
\end{array}
$$
and since each possible triple has distinct values, we can permute them to obtain:
$$ \bbox[lightyellow] { p(\text{3 consec.})={4/6} \cdot {1/6} \cdot {1/6} \cdot 6=1/9=24/216 }$$.
You can verify by direct counting that the values above are correct.
Best Answer
For four in a row, there are three possible lowest numbers. There are $4!$ ways to roll the required set of four numbers for each one, so the chance is $\frac {3 \cdot 4!}{6^4}=\frac {72}{1296}=\frac 1{18}$
Three in a row is harder. There are four possible lowest numbers of the run. Presumably we are prohibited from having a run of four. The fourth die can then match one of the three we have. There are three ways to choose which one, then $\frac {4!}2$ ways to order the throw for $4 \cdot 3 \cdot 12=144$ rolls. We can also have four distinct numbers with three in a row. Three are six rolls that satisfy this, $1235, 1236,2346,1345, 1456,2456$. These six possibilities can be arranged in $4!=24$ ways for a total of $144$. The total probability is then $\frac {144+144}{1296}=\frac 29$