Let the first term be \(a_1\) and the common difference be \(d\).

Using \(a_n=a_1+(n-1)d\), we get:

\[\left\{\begin{aligned}a_1+2d=6\\ a_1+4d=12\end{aligned}\right.\]

Subtract the first equation from the second:

\[2d=6\]

Hence, \(d=3\).

Then \(a_1=6-2\cdot 3=0\).

The sum of the first six terms is:

\[S_6=\frac{6}{2}\bigl(2a_1+5d\bigr)=3(0+15)=45\]

Answer: 45.

Let the first term be \(b_1\) and the common ratio be \(q\).

Using \(b_n=b_1q^{n-1}\), we have:

\[\left\{\begin{aligned}b_1q^2=60\\ b_1q^3=30\end{aligned}\right.\]

Divide the second equation by the first:

\[q=\frac{30}{60}=\frac{1}{2}\]

Then \(b_1\cdot \left(\frac12\right)^2=60\), so \(b_1=240\).

The first five terms are \(240,120,60,30,15\).

Therefore,

\[S_5=240+120+60+30+15=465\]

Answer: 465.

Using \(a_n=a_1+(n-1)d\):

\[a_5=a_1+4d=12\]

Substitute \(d=4\):

\[a_1+16=12\]

Hence, \(a_1=-4\).

Now find the sum of the first seven terms:

\[S_7=\frac{7}{2}\bigl(2a_1+6d\bigr)=\frac{7}{2}(-8+24)=\frac{7}{2}\cdot 16=56\]

Answer: 56.

Using \(b_2=b_1q\), we have:

\[b_1\cdot 2=4\]

Hence, \(b_1=2\).

The sum of the first five terms is:

\[S_5=b_1\frac{q^5-1}{q-1}=2\cdot\frac{2^5-1}{2-1}=2\cdot 31=62\]

Answer: 62.

For an arithmetic progression, \(a_n=a_1+(n-1)d\).

Then:

\[a_5-a_3=(a_1+4d)-(a_1+2d)=2d\]

Given that \(2d=6\), we get \(d=3\).

Now find the tenth term:

\[a_{10}=a_1+9d=20+9\cdot 3=47\]

Answer: 47.