From $5$ apples, $4$ mangoes and $3$ bananas, in how many ways we can select at least two fruits of each variety? 
From $5$ apples, $4$ mangoes and $3$ bananas, in how many ways we can select at least two fruits of each variety if fruits of same species are different?

My attempts:
$$\underbrace{\bigg({5\choose 2}+{5\choose 3}+{5\choose 4}+{5\choose 5}\bigg)}_{\text{appples}}\cdot\underbrace{\bigg({4\choose 2}+{4\choose 3}+{4\choose 4}\bigg)}_\text{mangoes}\cdot\underbrace{\bigg({3\choose 2}+{3\choose 3}\bigg)}_\text{bananas}=1144$$
But answer is: $2^{12}-4$. Please help.

Your solution is correct.  We can confirm it with another approach.
The number of subsets of a set with $n$ elements is $2^n$ since each element is either included in a subset or it is not.  
Since there are $5$ apples, there are $2^5$ subsets of apples.  Of these, $\binom{5}{0} + \binom{5}{1} = 1 + 5 = 6$ have fewer than two apples. Thus, $2^5 - 6 = 32 - 6 = 26$ have at least two apples. 
Since there are four mangoes, there are $2^4$ subsets of mangoes.  Of these, $\binom{4}{0} + \binom{4}{1} = 1 + 4 = 5$ have fewer than two mangoes.  Thus, $2^4 - 5 = 16 - 5 = 11$ have at least two mangoes.
Since there are three bananas, there are $2^3$ subsets of bananas.  Of these, $\binom{3}{0} + \binom{3}{1} = 1 + 3 = 4$ have fewer than two bananas.  Thus, $2^3 - 4 = 8 - 4 = 4$ have at least two bananas.
Hence, the number of ways of selecting at least two fruits of each variety is 
$$26 \cdot 11 \cdot 4 = 1144$$
as you found.
