How $\Delta f=\frac{1}{2T}$ satifies the orthogonality condition?
You're absolutely right, it must be $\Delta f = 1/T$, since this is the base frequency of the fourier transform of periodic functions with period $T$.

Otherwise, it's clear for $m=3$, for example, that the exponential would become 

$g(t)=\exp\left(j \frac{2\pi}{2T}\cdot 3 \cdot t\right)$

So it does not obey periodic boundary conditions at the right boundary, i.e. at $t=k\cdot T$ because

$g(T)=\exp(3\pi j)=-1 \not= 1 = g(0)$

So the problem statement in the book is wrong, but seems to come to the right result with either wrong calculus or by already knowing the right result. Maybe the author confused $\Delta f$ with the Nyquist frequency or something. Or it was just a typo in the book.

Remark: the sinc function has zeroes for all integers $n,m$ except where $n=m$, which is kind of a pitfall of $sinc(x)$, because the limit for $x\to 0$ is one, not zero as the occurrence of the sine in the numerator alone would imply. So the orthogonality condition that you have mentioned in the question could be extended to an orthonormality condition

$$&lt;s_{ml}(t),s_{nl}(t)&gt;=\delta_{mn}$$

which actually also holds for $n=m$, where it presents the normalizability of the basis functions (requiring appropriate normalization factors, of course).