1. basic calculations

Q. In the following rectangle sheet of 1X10, two numbers are shown. Fill the remaining rectangles by real numbers such that the sum of the numbers in any three neighbouring squares is a constant and the sum of all numbers is 210.

26

12

2. $\begin{array}{|c|} \hline
a\\ \hline
b\\ \hline
26\\ \hline
c\\ \hline
d\\ \hline
e\\ \hline
12\\ \hline
f\\ \hline
g\\ \hline
h\\ \hline
\end{array}$

From what you said;
a + b + 26 = k
b + 26 + c = k
26 + c + d = k
c + d + e = k
d + e + 12 = k
e + 12 + f = k
12 + f + g = k
f + g + h = k

You can start by using
26 + c + d = k
c + d + e = k
d + e + 12 = k

c + d = k - e

Hence
26 + k - e = k

26 - e = 0

And
d + e = k - c

Hence
k - c + 12 = k
12 - c = 0

Solve for c and e and the rest should come in easily.

3. My approach is basically the same as Unknown008's.

Here's what I said:

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline
a & b & 26 & c & d & e & 12 & f & g & h \\ \hline \end{array}$

And we have 9 equations:

$\begin{array}{cccccccccccccccccccc}
a &+& b &&&&&&&&&&&&& +& 26 &=& k & [1] \\
&& b &+& c &&&&&&&&&&& +& 26 &=& k & [2] \\
&&&& c &+& d &&&&&&&&& + & 26 &=& k & [3] \\
&&&& c &+& d &+& e &&&&&&&&& =& k & [4] \\
&&&&&& d &+& e &&&&&&& +& 12 &=& k & [5] \\
&&&&&&&& e &+& f &&&&& +& 12 &=& k & [6] \\
&&&&&&&&&& f &+& g &&& +& 12 &=& k & [7] \\
&&&&&&&&&& f &+& g &+& h &&& = & k & [8] \\ \\[-2mm]
a &+& b &+& c &+& d &+& e &+& f &+& g &+& h &+& 38 &=& 210 & [9]
\end{array}$

Subtract [4] - [3]: . $e - 26 \:=\:0 \quad\Rightarrow\quad e \:=\:26$

Subtract [4] - [5]: . $c - 12 \:=\:0 \quad\Rightarrow\quad c \:=\:12$

Subtract [8] - [7]: . $h-12 \:=\:0 \quad\Rightarrow\quad h \:=\:12$

. . . and we're on our way!