1.26: Please show your solutions.  Otherwise how do we find out where you made a mistake? :-) There is always one thing you can do to check your answer.  Plug in the value of $x$ back to the equation.  Does it work?  If it does, then you found a solution.  If not, then you made a mistake somewhere.

1.31: The question is: Solve the inequality $$\dfrac{1}{\log_2(x-1)} < \dfrac{1}{\log_2\sqrt{x+1}}.$$

The first thing to do in this kind of questions (equations or inequalities involving logarithm, fractions, and radicals) is to check the domains.  For what values of $x$ is the inequality defined?  In this question, we need to check the following: (1) denominators cannot be $0$; (2) The quantities to take logarithm should be positive; (3) The quantity inside square root should be nonnegative.  Do this before trying to solve anything.

So we have: (1) $\log_2(x-1)\neq 0$ and $\log_2\sqrt{x+1}\neq 0$; (2) $x-1 > 0$ and $\sqrt{x+1} > 0$; (3) $x+1\geq 0$.  From (1), $x\neq 2$ and $x\neq 0$.  From (2), $x > 1$ and $x > -1$, which means $x > 1$.  Here (3) is not a problem any more because (2)'s solution has already taken care of it.

Now, it is time to actually solve the inequality: from the above analysis of the domain of definition, $x\neq 2$ and $x > 1$, so we need to consider two intervals: $1 < x < 2$ and $x > 2$. 

If $1 < x < 2$, $0 < x-1 < 1$, so $\log_2(x-1) < 0$.  We also have $2 < x+1 < 3$, and thus $\sqrt{x+1} > \sqrt{2} > 1$, which means $\log_2\sqrt{x+1} > 0$.  The left hand side is negative and the right hand side is positive, so the inequality is always true in this interval (reciprocals do not change this result).  Therefore $1 < x < 2$ is a solution.

If $x > 2$, both sides are positive, then your solution applies.  At the end, you got $x > 3$ or $x < 0$, but since we are discussing within the constraint $x > 2$, the part of solution $x < 0$ is extraneous, therefore we only have $x > 3$.

Combining both cases, the full solution is the union of intervals $(1,2)\cup(3,+\infty)$.