$\mathbb{Z}_2\oplus\mathbb{Z}_9\oplus\mathbb{Z}_{35}$
Elementary Divisors: $2,3^2,5,7$.
Invariant Factors: $630$.
$\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$
Elementary Divisors: $2,2,2^3,2^3,3,5^2,5^3,7,7^2,13$.
Invariant Factors: $2,2,1400,1911000$.
Elementary divisors are fairly self-explanatory. You break each $n$ in the $\mathbb{Z}_n$ pieces into its unique prime factorization, $n=p_1^{r_1}\cdots p_m^{r_m}$, where the $p_i$ are distinct primes and the $r_i >0$. We then rewrite $\mathbb{Z}_n$ as:
$$\mathbb{Z}_n \cong \mathbb{Z}_{p_1^{r_1}}\oplus\mathbb{Z}_{p_2^{r_2}}\oplus\cdots\oplus\mathbb{Z}_{p_m^{r_m}}.$$
It is important to note that we cannot break the $\mathbb{Z}_9$ into $\mathbb{Z}_3\oplus\mathbb{Z}_3$ which is why the elementary divisors for the first group are $2,3^2,5,7$. For the invariant factors for the first group, see the example for the second group:
For the second group we see
$$\mathbb{Z}_{26}\cong\mathbb{Z}_{13}\oplus\mathbb{Z}_2 \text{ and } \mathbb{Z}_{42}\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_7$$
and so
$$\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\cong\mathbb{Z}_{13}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_7.$$
So in the expression
$$\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$$
we can replace the $\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}$ with
$$\mathbb{Z}_{13}\oplus\mathbb{Z}_2\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_3\oplus\mathbb{Z}_7.$$
The fact I have been using is that
$$\mathbb{Z}_{mn}\cong\mathbb{Z}_m\oplus\mathbb{Z}_n$$
whenever $\gcd(m,n)=1$.
We continue in this way until we have rewritten
$$H=\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$$
in the form
$$\mathbb{Z}_{p_1^{r_1}}\oplus\mathbb{Z}_{p_2^{r_2}}\cdots\oplus\mathbb{Z}_{p_m^{r_m}}$$
where all the $p_i$ are primes and the $r_i>0$. Doing so, we get that
$$H\cong \mathbb{Z}_{2}\oplus\mathbb{Z}_{13}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{3}\oplus\mathbb{Z}_{7}\oplus\mathbb{Z}_{7^2}\oplus\mathbb{Z}_{2^3}\oplus\mathbb{Z}_{5^2}\oplus\mathbb{Z}_{5^3}\oplus\mathbb{Z}_{2^3}.$$
For invariant factors, there is a trick once you know the elementary divisors. I will do the invariant factors for $\mathbb{Z}_{26}\oplus\mathbb{Z}_{42}\oplus\mathbb{Z}_{49}\oplus\mathbb{Z}_{200}\oplus\mathbb{Z}_{1000}$ now. I make this table out of the elementary divisors:
$$\begin{array}{c|c|c|c|c}
p=2 & p=3 & p=5 & p=7 & p=13 \\
\hline
2^3 & 3 & 5^3 & 7^2 & 13\\
2^3 &1&5^2&7&1\\
2&1&1&1&1\\
2&1&1&1&1
\end{array}$$
The first row I label each of the primes that appear as elementary divisors. In the next row I write the highest power of the given prime in each column. In the next row I write the elementary divisors of the next powers that I haven't used. I put a $1$ in a column when I have already recorded all of the elementary divisors of a given prime. It may be helpful to write down the elementary divisors and record them in a table on paper and compare to see that you understand how to do this. Next, take the product of the numbers in each row, and this will yield your invariant factors. In this case we get the invariant factors are:
$$2\cdot1\cdot1\cdot1\cdot1,2\cdot1\cdot1\cdot1\cdot1,2^3\cdot1\cdot5^2\cdot7\cdot1,2^3\cdot3\cdot5^3\cdot7^2\cdot13,$$
that is, the invariant factors are
$$2,2,1400,1911000.$$
Your last $$\mathbb Z_4\times \mathbb Z_2\times \mathbb Z_6\times \mathbb Z_6$$ has order 288. I assume that instead of $\mathbb Z_4$ you meant $\mathbb Z_2$.
Also, $$\mathbb Z_4\times \mathbb Z_{36}\not \cong \mathbb Z_{144}$$ since $\gcd(4, 36) = 4$.
I make use of the fact that $$\mathbb Z_{mn} \cong \mathbb Z_m \times \mathbb Z_n\;\;\text{if and only if}\;\;\gcd(m, n) = 1$$
$$\mathbb{Z}_4 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6 \not \cong \mathbb{Z}_{6} \oplus \mathbb{Z}_{24}$$
because $\gcd(4, 6) = 2 \neq 1$.
$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_6 \not\cong \mathbb{Z}_{12} \oplus \mathbb{Z}_{12}$$ because $\gcd(2, 6) = 2 \neq 1$.
Also, $3\mid 144,\;6\mid 144,\; 8\mid 144,\; 16\mid 144$...
$$\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_{36} \not \cong \mathbb{Z}_2 \oplus \mathbb{Z}_{72}$$ since $\gcd(2, 36) = 2\neq 1$.
The prime factorization of $144$ is given by: $$144 = 2^43^2$$
Best Answer
$\mathbb Z_2\oplus\mathbb Z_{12}\oplus\mathbb Z_{180}$ is right.
Your notes must be wrong because if the invariant factors were $\{2,2,6,6,30\}$ then there wouldn't be an element of order $36$ but $\mathbb Z_6\oplus\mathbb Z_{20}\oplus\mathbb Z_{36}$ has an element of order $36$ coming from $\mathbb Z_{36}$. This also gives elements of order $4$, $9$, $12$, which are not in $\{2,2,6,6,30\}$.