Is the language of words with equal number of 010s and 101s as substrings regular?
No, the language of words containing same number of 010s and 101s is not regular.

Consider $x_n=(0100)^n$ and $y_n=(1011)^n$. Then $x_iy_j$ contains $i$ number of 010s and $j$ number of $101s$. That is, $x_iy_j$ is in the language if and only if $i=j$. 

So $x_i$ represents a distinct Nerode-Myhill equivalence classes for all $i$. Since there are infinitely many such $x_i$, there are infinitely many Myhill-Nerode equivalence classes. By Myhill–Nerode theorem, the language is not regular.



Here are several related exercises. Exercise 2 answers the more general situation in the question. Exercise 3 is a further generalization. 

Let $\Sigma=\{0,1\}$.

Exercise 1. The language of words over alphabet $\Sigma$ containing the same number of 01s and 10s is regular. The language of words over alphabet $\{0,1, 2\}$ containing the same number of 01s and 10s is not regular.

Exercise 2. Let nonempty $w\in \Sigma^*$. Let $\overline w$ be the one's complement of $w$. For example, $\overline{010010111}=101101000$. Let $E_w$ be the language of words over $\Sigma$ containing the same number of $w$s and $\overline w$s. Show that 
$$ E_w\text{ is regular} \iff w=01\text{ or }w=10$$
(It may take a while to do this exercise.)

Exercise 3. Let $u$ and $v$ be two different non-empty words in $\Sigma^*$. Let $E_{u,v}$ be the language of words over $\Sigma$ containing the same number of $u$s and $v$s. Let $N_{u,v}$ be the language of words over $\Sigma$ containing no $u$ nor $v$ as a subword. Show that 
$$ L_{u,v}\text{ is regular } \iff 
\{u,v\}=\{01,10\}\text{ or } E_{u,v} = N_{u,v}.$$
(It may take a while to do this exercise.)