What is the general solution to equations of form $f(k\cdot x)-lf(x)=1, k,l \in \Bbb N$ ? Prove that $\log_{10} x$ is the solution when $k = 10,l =1$.

For the first question, $\log_{10}$ is not the only solution: let $g$ be any function defined on $[1, 10)$. Define $$f(x) = k + g(x \cdot 10 ^{-k})$$ when $10^k \le x \lt 10^{k+1}$. Then $f$ is a solution of the functional equation. Note that $k = \lfloor\log_{10}(x)\rfloor$

For the second equation, you can use a similar definition, I suggest $$f(x) = 5^k \left( g(x \cdot 10 ^{-k}) + \frac{1}{4}\right) - \frac{1}{4}$$

As Gribouillis points out, there are no unique solutions to equations like this. The equation only connects the values of $f(x)$ for values of $x$ that differ by factor ten, so it says nothing about $f(x)$ and $f(y)$ when $x/y$ is not an integer power of ten.

However, there is a way to find a "nice" solution to the kind of equation. If you want the general solution instead of pretty ones, this is not your answer. By pretty I mean something that satisfies a natural generalization of your condition.

Consider the second problem. Take any $x\in(0,\infty)$ and $n\in\mathbb N$. We have $$ \begin{split} f(10^nx) &= 1+5f(10^{n-1}x) \\&= 1+5(1+5f(10^{n-2}x)) \\&= (1+5)+5^2f(10^{n-2}x) \\&= \dots \\&= (1+5+25+\dots+5^{n-1})+5^nf(x) \\&= \frac14(5^n-1)+5^nf(x) \\&= 5^n(\frac14+f(x))-\frac14 . \end{split} $$ This formula can be used to find the general solution if you want to. To find a pretty solution, we add the additional restricting assumption that this holds for all values of $n\in\mathbb R$. For $x=1$ and $f(1)=a-1/4$ (the shift is for convenience) this leads to $$ f(y) = f(10^{\log_{10}y}) = %5^{\log_{10}y}(\frac14+a)-\frac14 5^{\log_{10}y}a-\frac14 . $$ It is easy to check that this indeed satisfies $f(10^tx) %=5^{t+\log_{10}x}a-\frac14=5^t5^{\log_{10}x}a-\frac14 =5^t(f(x)+1/4)-1/4$ for all $x>0$ and $t\in\mathbb R$.

You are free to choose the parameter $a$. For $a=0$ you get the constant function $f(x)=-1/4$.

Similarly, for the first problem you would get $f(y)=a+\log_{10}y$ with this method.