Here are some additional notes that the previous answerer did not mention.
#4: \(\sqrt{a}\) is irrational unless the radicand (the number inside the radical) is a perfect square (1,4,9,16,25, etc.). You can prove this by contradiction.
#5: We may not necessarily know an accurate decimal expansion of \(\sqrt{11}\), but you probably know that \(\sqrt{9}=3\) and \(\sqrt{16}=4\). This means that \(3<\sqrt{11}<4\). The only answer choice that corresponds to this restriction is D
#6: There's actually a trick we can use for this. It's quite clever, and it does not require converting into a decimal. Here it is again:
\(\frac{\textcolor{red}{3}}{\textcolor{blue}{4}}?\frac{\textcolor{blue}{4}}{\textcolor{red}{7}}\)
Multiply the numbers in blue and multiply the numbers in red. \(\textcolor{red}{3*7=21}\) and \(\textcolor{blue}{4*4=16}\). Since 3*7>4*4, this means that 3/4 is the larger fraction. Easy, isn't it?